Trigonometric identities

AS · P1 · 10 min

Paper 1 uses just two trigonometric identities, tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta} and sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1, but it uses them hard: to prove other identities, to simplify expressions, to find exact values, and above all to turn an equation that mixes functions into one you can solve. A "prove the identity" question worth 3 or 4 marks appears on most papers, often followed by "hence solve". This note covers the identities and how to prove things with them; solving the resulting equations is in Solving trigonometric equations.

Identities and equations

An equation such as sin⁡θ=12\sin\theta = \tfrac{1}{2} is true only for particular values of θ\theta, which you find by solving. An identity is true for every value of θ\theta for which both sides are defined. The symbol ≡\equiv ("is identically equal to") marks an identity.

That difference controls how you work. You solve an equation by doing the same thing to both sides. You prove an identity by transforming one side, step by step, until it becomes the other side.

The two identities

Tangent as a quotient

On the unit circle the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and tan⁡θ\tan\theta is the gradient of the radius to that point:

tan⁡θ≡sin⁡θcos⁡θ(cos⁡θ≠0)\tan\theta \equiv \frac{\sin\theta}{\cos\theta} \qquad (\cos\theta \neq 0)

The Pythagorean identity

The point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) lies on the circle x2+y2=1x^2 + y^2 = 1. Substituting gives

cos⁡2θ+sin⁡2θ≡1\cos^2\theta + \sin^2\theta \equiv 1

This is Pythagoras's theorem in the right-angled triangle with hypotenuse 11, and it holds for every angle. Here sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2.

y = sin(x)^2 y = cos(x)^2 y = 1

The graphs of y=sin⁡2xy = \sin^2 x and y=cos⁡2xy = \cos^2 x are mirror images about y=12y = \tfrac{1}{2}: wherever one is high the other is low, and at every xx their heights add up to exactly 11.

Key result
tan⁡θ≡sin⁡θcos⁡θ,sin⁡2θ+cos⁡2θ≡1\tan\theta \equiv \frac{\sin\theta}{\cos\theta}, \qquad \sin^2\theta + \cos^2\theta \equiv 1

Rearranged forms you will use constantly:

sin⁡2θ≡1−cos⁡2θ,cos⁡2θ≡1−sin⁡2θ\sin^2\theta \equiv 1 - \cos^2\theta, \qquad \cos^2\theta \equiv 1 - \sin^2\theta

Dividing sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1 by cos⁡2θ\cos^2\theta gives a third useful form:

tan⁡2θ+1≡1cos⁡2θ\tan^2\theta + 1 \equiv \frac{1}{\cos^2\theta}

The last form is just a consequence; in Paper 3 it is written with sec⁡θ\sec\theta. On Paper 1 derive it when you need it, as above.

Proving identities

Proving an identity
  1. Start with the more complicated side (usually the one with fractions, or with tan⁡\tan).
  2. Rewrite tan⁡θ\tan\theta as sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta} if it helps to have only sines and cosines.
  3. Combine fractions over a common denominator, or split a fraction into parts.
  4. Look for sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta (replace by 11), 1−sin⁡2θ1 - \sin^2\theta (replace by cos⁡2θ\cos^2\theta) or 1−cos⁡2θ1 - \cos^2\theta (replace by sin⁡2θ\sin^2\theta).
  5. Factorise, including differences of two squares: 1−cos⁡2θ=(1−cos⁡θ)(1+cos⁡θ)1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta) and sin⁡4θ−cos⁡4θ=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta).
  6. Stop when you reach the other side exactly, and write a concluding line: "== RHS".

A useful trick for fractions with 1±cos⁡θ1 \pm \cos\theta or 1±sin⁡θ1 \pm \sin\theta: multiply top and bottom by the "conjugate" (for example 1+cos⁡θ1 + \cos\theta), so the denominator becomes 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.

If you cannot see how to start, try working on both sides separately until they meet in the middle. This is a valid proof provided every step is reversible and you say clearly that both sides equal the same expression.

Using identities to find values

One ratio from another, algebraically

The triangle method in Exact values and angles of any size is usually quickest, but the identity works too: if sin⁡θ=35\sin\theta = \tfrac{3}{5} then cos⁡2θ=1−925=1625\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}, so cos⁡θ=±45\cos\theta = \pm\tfrac{4}{5}, and the quadrant chooses the sign.

Expressions in sin⁡θ\sin\theta and cos⁡θ\cos\theta given tan⁡θ\tan\theta

An expression whose top and bottom are both of the form asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta can be found directly from tan⁡θ\tan\theta: divide every term by cos⁡θ\cos\theta.

Rewriting equations

The most important use. An equation such as 2cos⁡2θ+sin⁡θ=12\cos^2\theta + \sin\theta = 1 contains two different functions. Replacing cos⁡2θ\cos^2\theta by 1−sin⁡2θ1 - \sin^2\theta gives a quadratic in sin⁡θ\sin\theta alone, which can then be solved (see Equations that are quadratic in a function of x). Cambridge usually asks you to "show that the equation can be written as ..." first.

Worked examples

Combining fractions

Prove the identity tan⁡θ+1tan⁡θ≡1sin⁡θcos⁡θ\tan\theta + \dfrac{1}{\tan\theta} \equiv \dfrac{1}{\sin\theta\cos\theta}.

SolutionLHS=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ=RHS\begin{aligned} \text{LHS} &= \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \\ &= \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \\ &= \frac{1}{\sin\theta\cos\theta} = \text{RHS} \end{aligned}
Multiplying by the conjugate

Prove that 1−cos⁡θsin⁡θ≡sin⁡θ1+cos⁡θ\dfrac{1 - \cos\theta}{\sin\theta} \equiv \dfrac{\sin\theta}{1 + \cos\theta}.

Solution

Multiply the top and bottom of the left-hand side by 1+cos⁡θ1 + \cos\theta:

LHS=(1−cos⁡θ)(1+cos⁡θ)sin⁡θ(1+cos⁡θ)=1−cos⁡2θsin⁡θ(1+cos⁡θ)=sin⁡2θsin⁡θ(1+cos⁡θ)=sin⁡θ1+cos⁡θ=RHS\begin{aligned} \text{LHS} &= \frac{(1 - \cos\theta)(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} \\ &= \frac{1 - \cos^2\theta}{\sin\theta(1 + \cos\theta)} \\ &= \frac{\sin^2\theta}{\sin\theta(1 + \cos\theta)} \\ &= \frac{\sin\theta}{1 + \cos\theta} = \text{RHS} \end{aligned}
A difference of two squares

Prove that sin⁡4θ−cos⁡4θ≡1−2cos⁡2θ\sin^4\theta - \cos^4\theta \equiv 1 - 2\cos^2\theta.

SolutionLHS=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)=sin⁡2θ−cos⁡2θ=(1−cos⁡2θ)−cos⁡2θ=1−2cos⁡2θ=RHS\begin{aligned} \text{LHS} &= \left(\sin^2\theta - \cos^2\theta\right)\left(\sin^2\theta + \cos^2\theta\right) \\ &= \sin^2\theta - \cos^2\theta \\ &= \left(1 - \cos^2\theta\right) - \cos^2\theta \\ &= 1 - 2\cos^2\theta = \text{RHS} \end{aligned}
A Cambridge-style proof with tangent

Prove the identity sin⁡θ−tan⁡θsin⁡θ+tan⁡θ≡cos⁡θ−1cos⁡θ+1\dfrac{\sin\theta - \tan\theta}{\sin\theta + \tan\theta} \equiv \dfrac{\cos\theta - 1}{\cos\theta + 1}.

Solution

Replace tan⁡θ\tan\theta and multiply the top and bottom by cos⁡θ\cos\theta:

LHS=sin⁡θ−sin⁡θcos⁡θsin⁡θ+sin⁡θcos⁡θ=sin⁡θcos⁡θ−sin⁡θsin⁡θcos⁡θ+sin⁡θ=sin⁡θ(cos⁡θ−1)sin⁡θ(cos⁡θ+1)=cos⁡θ−1cos⁡θ+1=RHS\begin{aligned} \text{LHS} &= \frac{\sin\theta - \dfrac{\sin\theta}{\cos\theta}}{\sin\theta + \dfrac{\sin\theta}{\cos\theta}} \\ &= \frac{\sin\theta\cos\theta - \sin\theta}{\sin\theta\cos\theta + \sin\theta} \\ &= \frac{\sin\theta(\cos\theta - 1)}{\sin\theta(\cos\theta + 1)} \\ &= \frac{\cos\theta - 1}{\cos\theta + 1} = \text{RHS} \end{aligned}
Values from the tangent

Given that tan⁡θ=2\tan\theta = 2, find the exact value of sin⁡θ+3cos⁡θ2sin⁡θ−cos⁡θ\dfrac{\sin\theta + 3\cos\theta}{2\sin\theta - \cos\theta}.

Solution

Divide every term, top and bottom, by cos⁡θ\cos\theta:

sin⁡θ+3cos⁡θ2sin⁡θ−cos⁡θ=tan⁡θ+32tan⁡θ−1=2+34−1=53\frac{\sin\theta + 3\cos\theta}{2\sin\theta - \cos\theta} = \frac{\tan\theta + 3}{2\tan\theta - 1} = \frac{2 + 3}{4 - 1} = \frac{5}{3}

The quadrant of θ\theta does not matter: the answer is the same whether θ\theta is in the first or third quadrant.

Rewriting an equation

Show that the equation 2tan⁡θsin⁡θ=32\tan\theta\sin\theta = 3 can be written as 2cos⁡2θ+3cos⁡θ−2=02\cos^2\theta + 3\cos\theta - 2 = 0, and hence find the values of θ\theta for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.

Solution2×sin⁡θcos⁡θ×sin⁡θ=32sin⁡2θ=3cos⁡θ2(1−cos⁡2θ)=3cos⁡θ2cos⁡2θ+3cos⁡θ−2=0\begin{aligned} 2 \times \frac{\sin\theta}{\cos\theta} \times \sin\theta &= 3 \\ 2\sin^2\theta &= 3\cos\theta \\ 2\left(1 - \cos^2\theta\right) &= 3\cos\theta \\ 2\cos^2\theta + 3\cos\theta - 2 &= 0 \end{aligned}

Factorise: (2cos⁡θ−1)(cos⁡θ+2)=0(2\cos\theta - 1)(\cos\theta + 2) = 0. cos⁡θ=−2\cos\theta = -2 is impossible, so cos⁡θ=12\cos\theta = \tfrac{1}{2}, giving θ=60∘\theta = 60^\circ or 360∘−60∘=300∘360^\circ - 60^\circ = 300^\circ. (The method for finding all solutions is in Solving trigonometric equations.)

An identity with two fractions

Prove that sin⁡θ1−cos⁡θ−1sin⁡θ≡1tan⁡θ\dfrac{\sin\theta}{1 - \cos\theta} - \dfrac{1}{\sin\theta} \equiv \dfrac{1}{\tan\theta}.

SolutionLHS=sin⁡2θ−(1−cos⁡θ)sin⁡θ(1−cos⁡θ)=(1−cos⁡2θ)−1+cos⁡θsin⁡θ(1−cos⁡θ)=cos⁡θ−cos⁡2θsin⁡θ(1−cos⁡θ)=cos⁡θ(1−cos⁡θ)sin⁡θ(1−cos⁡θ)=cos⁡θsin⁡θ=1tan⁡θ=RHS\begin{aligned} \text{LHS} &= \frac{\sin^2\theta - (1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} \\ &= \frac{\left(1 - \cos^2\theta\right) - 1 + \cos\theta}{\sin\theta(1 - \cos\theta)} \\ &= \frac{\cos\theta - \cos^2\theta}{\sin\theta(1 - \cos\theta)} \\ &= \frac{\cos\theta(1 - \cos\theta)}{\sin\theta(1 - \cos\theta)} \\ &= \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} = \text{RHS} \end{aligned}
Eliminating the angle

A point moves so that its coordinates are x=3cos⁡θx = 3\cos\theta and y=2sin⁡θy = 2\sin\theta. Find an equation connecting xx and yy that does not involve θ\theta.

Solution

cos⁡θ=x3\cos\theta = \tfrac{x}{3} and sin⁡θ=y2\sin\theta = \tfrac{y}{2}. Using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1

(This curve is an ellipse; you do not need to know that for Paper 1.)

Watch out

Treating an identity like an equation. Do not start with "LHS == RHS" and move terms across: that assumes what you are trying to prove. Work on one side only.

sin⁡2θ\sin^2\theta versus sin⁡θ2\sin\theta^2. sin⁡2θ=(sin⁡θ)2\sin^2\theta = (\sin\theta)^2. Writing sin⁡θ2\sin\theta^2 suggests the sine of θ2\theta^2.

False "identities". sin⁡θ+cos⁡θ\sin\theta + \cos\theta is not 11, sin⁡2θ\sin 2\theta is not 2sin⁡θ2\sin\theta, and sin⁡(A+B)\sin(A + B) is not sin⁡A+sin⁡B\sin A + \sin B.

Cancelling terms instead of factors. In sin⁡θ(cos⁡θ−1)sin⁡θ(cos⁡θ+1)\dfrac{\sin\theta(\cos\theta - 1)}{\sin\theta(\cos\theta + 1)} you may cancel sin⁡θ\sin\theta because it multiplies everything. In sin⁡θ−tan⁡θsin⁡θ+tan⁡θ\dfrac{\sin\theta - \tan\theta}{\sin\theta + \tan\theta} you may not cancel sin⁡θ\sin\theta.

Losing a sign with 1−cos⁡θ1 - \cos\theta. sin⁡2θ−(1−cos⁡θ)=sin⁡2θ−1+cos⁡θ\sin^2\theta - (1 - \cos\theta) = \sin^2\theta - 1 + \cos\theta. Bracket what you subtract.

Exam tip
  • "Prove" or "show that". Every line should follow from the one before, and the final line must match the target exactly. Examiners withhold the last mark for a proof that "fizzles out" with no conclusion.
  • Use ≡\equiv or == consistently. Writing "LHS =…== \ldots = RHS" is the clearest layout.
  • "Hence". After proving an identity, "hence solve" means replace the left side by the simpler right side, then solve. Not using the identity usually means a long or impossible route.
  • Show the identity substitution. Write 2(1−cos⁡2θ)2(1 - \cos^2\theta) before expanding, so the examiner sees which identity you used.
  • Values from tangent. Divide by cos⁡θ\cos\theta rather than finding sin⁡θ\sin\theta and cos⁡θ\cos\theta separately; it avoids sign errors.
Summary
  • An identity holds for all values; prove it by transforming one side into the other.
  • tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta} and sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1; also sin⁡2θ≡1−cos⁡2θ\sin^2\theta \equiv 1 - \cos^2\theta, cos⁡2θ≡1−sin⁡2θ\cos^2\theta \equiv 1 - \sin^2\theta, tan⁡2θ+1≡1cos⁡2θ\tan^2\theta + 1 \equiv \dfrac{1}{\cos^2\theta}.
  • Proof tactics: write tan in terms of sin and cos, combine fractions, use the Pythagorean identity, factorise (including differences of squares), multiply by 1±cos⁡θ1 \pm \cos\theta.
  • Given tan⁡θ\tan\theta, evaluate asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ\dfrac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} by dividing by cos⁡θ\cos\theta.
  • To solve an equation mixing sin⁡2\sin^2 and cos⁡\cos (or cos⁡2\cos^2 and sin⁡\sin), replace the square so only one function remains.
  • To eliminate θ\theta from expressions for sin⁡θ\sin\theta and cos⁡θ\cos\theta, use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

Practice questions

Question
  1. Prove that 1−sin⁡2θcos⁡θ≡cos⁡θ\dfrac{1 - \sin^2\theta}{\cos\theta} \equiv \cos\theta.
  2. Prove that cos⁡θ+sin⁡θtan⁡θ≡1cos⁡θ\cos\theta + \sin\theta\tan\theta \equiv \dfrac{1}{\cos\theta}.
  3. Prove that sin⁡4θ+sin⁡2θcos⁡2θ≡sin⁡2θ\sin^4\theta + \sin^2\theta\cos^2\theta \equiv \sin^2\theta.
  4. Prove that 11−cos⁡x+11+cos⁡x≡2sin⁡2x\dfrac{1}{1 - \cos x} + \dfrac{1}{1 + \cos x} \equiv \dfrac{2}{\sin^2 x}.
  5. Prove that tan⁡2θ−sin⁡2θ≡tan⁡2θsin⁡2θ\tan^2\theta - \sin^2\theta \equiv \tan^2\theta\sin^2\theta.
  6. Given that tan⁡θ=3\tan\theta = 3, find the exact value of 2sin⁡θ−cos⁡θsin⁡θ+cos⁡θ\dfrac{2\sin\theta - \cos\theta}{\sin\theta + \cos\theta}.
  7. Show that the equation 3tan⁡θ=2cos⁡θ3\tan\theta = 2\cos\theta can be written as 2sin⁡2θ+3sin⁡θ−2=02\sin^2\theta + 3\sin\theta - 2 = 0, and hence find the values of θ\theta for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
  8. Given that cos⁡θ=−23\cos\theta = -\tfrac{2}{3} and θ\theta is in the third quadrant, use the identity sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1 to find the exact values of sin⁡θ\sin\theta and tan⁡θ\tan\theta.
  9. A point has coordinates x=2+3cos⁡θx = 2 + 3\cos\theta, y=1−3sin⁡θy = 1 - 3\sin\theta. Show that, for all values of θ\theta, the point lies on a circle, and state its centre and radius.
  10. (a) Prove that 1+sin⁡θcos⁡θ+cos⁡θ1+sin⁡θ≡2cos⁡θ\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} \equiv \dfrac{2}{\cos\theta}. (b) Hence solve 1+sin⁡θcos⁡θ+cos⁡θ1+sin⁡θ=4\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = 4 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
Answers

  1. 1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ\dfrac{1 - \sin^2\theta}{\cos\theta} = \dfrac{\cos^2\theta}{\cos\theta} = \cos\theta.

  2. cos⁡θ+sin⁡θ⋅sin⁡θcos⁡θ=cos⁡2θ+sin⁡2θcos⁡θ=1cos⁡θ\cos\theta + \sin\theta \cdot \dfrac{\sin\theta}{\cos\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\cos\theta} = \dfrac{1}{\cos\theta}.

  3. sin⁡4θ+sin⁡2θcos⁡2θ=sin⁡2θ(sin⁡2θ+cos⁡2θ)=sin⁡2θ\sin^4\theta + \sin^2\theta\cos^2\theta = \sin^2\theta\left(\sin^2\theta + \cos^2\theta\right) = \sin^2\theta.

  4. (1+cos⁡x)+(1−cos⁡x)(1−cos⁡x)(1+cos⁡x)=21−cos⁡2x=2sin⁡2x\dfrac{(1 + \cos x) + (1 - \cos x)}{(1 - \cos x)(1 + \cos x)} = \dfrac{2}{1 - \cos^2 x} = \dfrac{2}{\sin^2 x}.

  5. tan⁡2θ−sin⁡2θ=sin⁡2θcos⁡2θ−sin⁡2θ=sin⁡2θ(1−cos⁡2θ)cos⁡2θ=sin⁡2θcos⁡2θ×sin⁡2θ=tan⁡2θsin⁡2θ\tan^2\theta - \sin^2\theta = \dfrac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta = \dfrac{\sin^2\theta\left(1 - \cos^2\theta\right)}{\cos^2\theta} = \dfrac{\sin^2\theta}{\cos^2\theta} \times \sin^2\theta = \tan^2\theta\sin^2\theta.

  6. Dividing by cos⁡θ\cos\theta: 2tan⁡θ−1tan⁡θ+1=6−13+1=54\dfrac{2\tan\theta - 1}{\tan\theta + 1} = \dfrac{6 - 1}{3 + 1} = \dfrac{5}{4}.

  7. 3sin⁡θcos⁡θ=2cos⁡θ\dfrac{3\sin\theta}{\cos\theta} = 2\cos\theta, so 3sin⁡θ=2cos⁡2θ=2−2sin⁡2θ3\sin\theta = 2\cos^2\theta = 2 - 2\sin^2\theta, giving 2sin⁡2θ+3sin⁡θ−2=02\sin^2\theta + 3\sin\theta - 2 = 0. Factorise: (2sin⁡θ−1)(sin⁡θ+2)=0(2\sin\theta - 1)(\sin\theta + 2) = 0. sin⁡θ=−2\sin\theta = -2 is impossible; sin⁡θ=12\sin\theta = \tfrac{1}{2} gives θ=30∘\theta = 30^\circ or 150∘150^\circ.

  8. sin⁡2θ=1−49=59\sin^2\theta = 1 - \tfrac{4}{9} = \tfrac{5}{9}, so sin⁡θ=±53\sin\theta = \pm\tfrac{\sqrt{5}}{3}. In the third quadrant sine is negative: sin⁡θ=−53\sin\theta = -\tfrac{\sqrt{5}}{3}. Then tan⁡θ=−5/3−2/3=52\tan\theta = \dfrac{-\sqrt{5}/3}{-2/3} = \dfrac{\sqrt{5}}{2}.

  9. cos⁡θ=x−23\cos\theta = \dfrac{x - 2}{3} and sin⁡θ=1−y3\sin\theta = \dfrac{1 - y}{3}. Then (x−2)29+(y−1)29=1\dfrac{(x - 2)^2}{9} + \dfrac{(y - 1)^2}{9} = 1, i.e. (x−2)2+(y−1)2=9(x - 2)^2 + (y - 1)^2 = 9: a circle with centre (2,1)(2, 1) and radius 33.

  10. (a) (1+sin⁡θ)2+cos⁡2θcos⁡θ(1+sin⁡θ)=1+2sin⁡θ+sin⁡2θ+cos⁡2θcos⁡θ(1+sin⁡θ)=2+2sin⁡θcos⁡θ(1+sin⁡θ)=2(1+sin⁡θ)cos⁡θ(1+sin⁡θ)=2cos⁡θ\dfrac{(1 + \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 + \sin\theta)} = \dfrac{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}{\cos\theta(1 + \sin\theta)} = \dfrac{2 + 2\sin\theta}{\cos\theta(1 + \sin\theta)} = \dfrac{2(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)} = \dfrac{2}{\cos\theta}. (b) 2cos⁡θ=4\dfrac{2}{\cos\theta} = 4, so cos⁡θ=12\cos\theta = \tfrac{1}{2} and θ=60∘\theta = 60^\circ or 300∘300^\circ. Neither makes cos⁡θ=0\cos\theta = 0 or 1+sin⁡θ=01 + \sin\theta = 0, so both are valid.

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