Equations with Trigonometric Identities | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Choose what needs to disappear

Use what the equation contains. A common factor suggests factorisation. Mixed sine and cosine squares suggest $\sin^2\theta+\cos^2\theta=1$ . Tangent with sine or cosine suggests replacing tangent by their ratio.

1. One trig functionChoose an identity or factorisation.
2. Algebraic rootsUse the polynomial solver when suitable.
3. Valid ratiosApply the range and denominator conditions.
4. All anglesKeep every branch inside the interval.

Each algebraic root still needs an angle check.

Before clearing a denominator, record where the original expression is undefined. Before dividing by a trig function, check its zero case.

Worked example

Solve $3\cos^2x-\sin x\cos x=0$ on $0^\circ\le x\le360^\circ$ .

\cos x(3\cos x-\sin x)=0

The first factor gives $x=90^\circ,270^\circ$ . For the second factor, cosine cannot be zero: that would also require sine to be zero, which never happens at the same angle. So division by cosine is now safe:

3\cos x=\sin x

\tan x=3

The other angles are 71.6° and 251.6°. Keep all four, including the two zero-cosine solutions.

When the quadratic contains a squared ratio

Worked example

Use an identity to solve $3\tan^2\theta+5\sin^2\theta=9$ for $0^\circ<\theta<270^\circ$ .

(9709/12 Feb/Mar 2025 Q7 [7 marks])

Part (a) asks for the expression in terms of sine. Write tangent squared as sine squared over cosine squared. Replace cosine squared by $1-\sin^2\theta$ :

3\tan^2\theta+5\sin^2\theta

=\frac{3\sin^2\theta}{1-\sin^2\theta}+5\sin^2\theta

Use the common denominator. The original tangent excludes $\cos\theta=0$ , so this denominator is non-zero.

Let $u=\sin^2\theta$ to keep the algebra short. Then $\sin^4\theta=u^2$ .

\frac{3u+5u(1-u)}{1-u}

=\frac{8u-5u^2}{1-u}

=\frac{8\sin^2\theta-5\sin^4\theta}{1-\sin^2\theta}

For part (b), set this equal to 9 and multiply by the denominator:

8u-5u^2=9(1-u)

Expand the bracket and bring all terms to one side:

5u^2-17u+9=0

The polynomial solver finds about 0.65597 and 2.74403. These are not convenient factors; the quadratic formula supplies exact working:

\begin{aligned}u&=\frac{17\pm\sqrt{17^2-180}}{10}\\ &=\frac{17\pm\sqrt{109}}{10}\end{aligned}

Since $u$ is sine squared, keep only $0\le u\le1$ . Taking the square root gives both signs:

\sin\theta=\pm\sqrt{\frac{17-\sqrt{109}}{10}}

Using the unrounded value, the acute reference angle is $54.088\ldots^\circ$ . Quadrants I, II and III are in the interval; IV is outside it.

Answer

\theta=54.1^\circ,\ 125.9^\circ,\ 234.1^\circ

None is an excluded tangent angle. Substitute the unrounded values into the original equation to check the result.

Your turn

Your turnMixed practice

Solve each equation for $0^\circ\le x<360^\circ$ .

(a) $2\sin^2x+3\cos x=3$ .

(b) $\sin x\cos x=\sin x$ .

Show worked answer

(a) Keep cosine by replacing sine squared:

2(1-\cos^2x)+3\cos x=3

2\cos^2x-3\cos x+1=0

The roots give $(2\cos x-1)(\cos x-1)=0$ . Thus cosine is 1/2 or 1.

Answer

x=0^\circ,\ 60^\circ,\ 300^\circ

(b) Factor: $\sin x(\cos x-1)=0$ . Sine zero gives 0° and 180°. Cosine 1 gives 0° again, not an extra solution.

Answer

x=0^\circ,\ 180^\circ

360° is excluded in both parts. Part (b) would lose 180° if you divided by sine at the start.

Choose what needs to disappear

1. One trig functionChoose an identity or factorisation.
2. Algebraic rootsUse the polynomial solver when suitable.
3. Valid ratiosApply the range and denominator conditions.
4. All anglesKeep every branch inside the interval.

Each algebraic root still needs an angle check.

Before clearing a denominator, record where the original expression is undefined. Before dividing by a trig function, check its zero case.

Worked example

Solve $3\cos^2x-\sin x\cos x=0$ on $0^\circ\le x\le360^\circ$ .

\cos x(3\cos x-\sin x)=0

3\cos x=\sin x

\tan x=3

The other angles are 71.6° and 251.6°. Keep all four, including the two zero-cosine solutions.

When the quadratic contains a squared ratio

Worked example

Use an identity to solve $3\tan^2\theta+5\sin^2\theta=9$ for $0^\circ<\theta<270^\circ$ .

(9709/12 Feb/Mar 2025 Q7 [7 marks])

Part (a) asks for the expression in terms of sine. Write tangent squared as sine squared over cosine squared. Replace cosine squared by $1-\sin^2\theta$ :

3\tan^2\theta+5\sin^2\theta

=\frac{3\sin^2\theta}{1-\sin^2\theta}+5\sin^2\theta

Use the common denominator. The original tangent excludes $\cos\theta=0$ , so this denominator is non-zero.

Let $u=\sin^2\theta$ to keep the algebra short. Then $\sin^4\theta=u^2$ .

\frac{3u+5u(1-u)}{1-u}

=\frac{8u-5u^2}{1-u}

=\frac{8\sin^2\theta-5\sin^4\theta}{1-\sin^2\theta}

For part (b), set this equal to 9 and multiply by the denominator:

8u-5u^2=9(1-u)

Expand the bracket and bring all terms to one side:

5u^2-17u+9=0

The polynomial solver finds about 0.65597 and 2.74403. These are not convenient factors; the quadratic formula supplies exact working:

\begin{aligned}u&=\frac{17\pm\sqrt{17^2-180}}{10}\\ &=\frac{17\pm\sqrt{109}}{10}\end{aligned}

Since $u$ is sine squared, keep only $0\le u\le1$ . Taking the square root gives both signs:

\sin\theta=\pm\sqrt{\frac{17-\sqrt{109}}{10}}

Using the unrounded value, the acute reference angle is $54.088\ldots^\circ$ . Quadrants I, II and III are in the interval; IV is outside it.

Answer

\theta=54.1^\circ,\ 125.9^\circ,\ 234.1^\circ

None is an excluded tangent angle. Substitute the unrounded values into the original equation to check the result.

Your turn

Your turnMixed practice

Solve each equation for $0^\circ\le x<360^\circ$ .

(a) $2\sin^2x+3\cos x=3$ .

(b) $\sin x\cos x=\sin x$ .

Show worked answer

(a) Keep cosine by replacing sine squared:

2(1-\cos^2x)+3\cos x=3

2\cos^2x-3\cos x+1=0

The roots give $(2\cos x-1)(\cos x-1)=0$ . Thus cosine is 1/2 or 1.

Answer

x=0^\circ,\ 60^\circ,\ 300^\circ

(b) Factor: $\sin x(\cos x-1)=0$ . Sine zero gives 0° and 180°. Cosine 1 gives 0° again, not an extra solution.

Answer

x=0^\circ,\ 180^\circ

360° is excluded in both parts. Part (b) would lose 180° if you divided by sine at the start.