Sec 4 A-Math Trigonometric identities and formulae worksheet with answers

A free Sec 4 A-Math worksheet on trigonometric identities and formulae: 7 questions, 40 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Trigonometric identities and formulae notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Trigonometric identities and formulae

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 4 A-Math worksheet: Trigonometric identities and formulae

Total: 40 marksSuggested time: 1 hour

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

Express

3\cos\theta + 5\sin\theta

in the form

R\cos(\theta - \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

. Find the exact value of

R

Answer[2]

(a)Given that

\sin A = \dfrac{4}{5}

and

A

is acute, find the exact value of

\cos A

Answer (a)[2]

(b)Given that

\sin A = -\dfrac{3}{5}

and

180^\circ < A < 270^\circ

, find the exact value of

\sec A

Answer (b)[2]

(a)Express

2\sin\theta + 3\cos\theta

in the form

R\sin(\theta + \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

. Find the exact value of

R

Answer (a)[2]

(b)Express

3\sin\theta + 4\cos\theta

in the form

R\sin(\theta + \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

. Hence solve

3\sin\theta + 4\cos\theta = -1

for

0^\circ \leq \theta \leq 360^\circ

Answer (b)[3]

(a)Solve

\sin 2x = 1.5\sin x

for

0^\circ \leq x \leq 360^\circ

Answer (a)[3]

(b)Solve

\sin 2x = 0.6\sin x

for

0^\circ \leq x \leq 360^\circ

Answer (b)[3]

Section B

(a)Show that the equation

2\sec^{2} x - \tan x - 3 = 0

can be written as

2\tan^{2} x - \tan x - 1 = 0

[2]

(b)Hence solve the equation

2\sec^{2} x - \tan x - 3 = 0

for

0^\circ \leq x \leq 360^\circ

. Give your answers correct to 1 decimal place.

Answer (b)[4]

(c)Hence state the number of solutions of the equation

2\sec^{2} 2x - \tan 2x - 3 = 0

for

0^\circ \leq x \leq 360^\circ

Answer (c)[1]

It is given that

\cos(2A - B) = 2\cos(2A + B)

(a)Show that

\cot 2A\cot B = 3

[3]

(b)Given that

B = 45^\circ

, find the exact value of

\tan 2A

Answer (b)[1]

(c)Hence solve the equation for

0^\circ < A < 180^\circ

, giving your answers correct to 1 decimal place.

Answer (c)[3]

It is given that

12\sin\theta + 5\cos\theta = R\sin(\theta + \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

(a)Find the exact value of

R

Answer (a)[2]

(b)Find the value of

\alpha

, correct to 2 decimal places.

Answer (b)[2]

(c)Find the value of

\theta

, where

0^\circ \leq \theta \leq 360^\circ

, for which

12\sin\theta + 5\cos\theta

is greatest.

Answer (c)[2]

(d)Solve

12\sin\theta + 5\cos\theta = -1

for

0^\circ \leq \theta \leq 360^\circ

Answer (d)[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= \sqrt{34}

[2]

R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha

R\cos\alpha = 3

R\sin\alpha = 5

R = \sqrt{3^{2} + 5^{2}}

= \sqrt{34}

Question 2

(a)Answer:

\cos A = \dfrac{3}{5}

[2]

\text{Adjacent} = \sqrt{5^{2} - 4^{2}}

= 3

A

is in quadrant 1, so

\cos A

is positive.

\cos A = \dfrac{3}{5}

(b)Answer:

\sec A = -\dfrac{5}{4}

[2]

\text{Adjacent} = \sqrt{5^{2} - 3^{2}}

= 4

A

is in quadrant 3, so

\sec A

is negative.

\sec A = -\dfrac{5}{4}

Question 3

(a)Answer:

= \sqrt{13}

[2]

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

R\cos\alpha = 2

R\sin\alpha = 3

R = \sqrt{2^{2} + 3^{2}}

= \sqrt{13}

(b)Answer:

\theta = 295.3^\circ

[3]

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

R\cos\alpha = 3

R\sin\alpha = 4

R = \sqrt{3^{2} + 4^{2}}

= 5

\tan\alpha = \dfrac{4}{3}

\alpha = 53.13^\circ

(2 d.p.)

3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13^\circ)

\sin(\theta + 53.13^\circ) = -\dfrac{1}{5}

= -0.2000\ldots

\beta = \sin^{-1}\left(0.2000\ldots\right)

\beta = 11.5^\circ

\sin u

is negative in quadrants 3 and 4.

u = 180^\circ + 11.5^\circ

= 191.5^\circ

u = 360^\circ - 11.5^\circ

= 348.5^\circ

\theta = 138.4^\circ

\theta = 295.3^\circ

Question 4

(a)Answer:

= 318.6^\circ

[3]

2\sin x\cos x - 1.5\sin x = 0

\sin x\Big(2\cos x - 1.5\Big) = 0

\sin x = 0

x = 0^\circ

x = 180^\circ

x = 360^\circ

\cos x = \dfrac{3}{4}

\beta = \cos^{-1}\left(0.7500\ldots\right)

\beta = 41.4^\circ

\cos x

is positive in quadrants 1 and 4.

x = 41.4^\circ

x = 360^\circ - 41.4^\circ

= 318.6^\circ

(b)Answer:

= 287.5^\circ

[3]

2\sin x\cos x - 0.6\sin x = 0

\sin x\Big(2\cos x - 0.6\Big) = 0

\sin x = 0

x = 0^\circ

x = 180^\circ

x = 360^\circ

\cos x = \dfrac{3}{10}

\beta = \cos^{-1}\left(0.3000\ldots\right)

\beta = 72.5^\circ

\cos x

is positive in quadrants 1 and 4.

x = 72.5^\circ

x = 360^\circ - 72.5^\circ

= 287.5^\circ

Question 5

(a)[2]

Using

\sec^{2} x = 1 + \tan^{2} x

2(1 + \tan^{2} x) - \tan x - 3 = 0

2\tan^{2} x - \tan x - 1 = 0

(shown)

(b)Answer:

x = 45.0^\circ, 153.4^\circ, 225.0^\circ, 333.4^\circ

[4]

2\tan^{2} x - \tan x - 1 = 0

\tan x = -\dfrac{1}{2}

\tan x = 1

x = 45.0^\circ, 153.4^\circ, 225.0^\circ, 333.4^\circ

(c)Answer:

8

[1]

Each solution for

2x

0^\circ \leq 2x \leq 720^\circ

gives one for

x

.There are

8

solutions.

Question 6

(a)[3]

\cos 2A\cos B + \sin 2A\sin B = 2(\cos 2A\cos B - \sin 2A\sin B)

\sin 2A\sin B + 2\sin 2A\sin B = 2\cos 2A\cos B - \cos 2A\cos B

(3)\sin 2A\sin B = (1)\cos 2A\cos B

\dfrac{\cos 2A\cos B}{\sin 2A\sin B} = 3

\cot 2A\cot B = 3

(shown)

(b)Answer:

\tan 2A = \dfrac{1}{3}

[1]

\cot 45^\circ = 1

, so

\cot 2A = 3

\tan 2A = \dfrac{1}{3}

(c)Answer:

A = 9.2^\circ, 99.2^\circ

[3]

\tan 2A = \dfrac{1}{3}

0^\circ < 2A < 360^\circ

Basic angle

= 18.43^\circ

2A = 18.43^\circ, 198.43^\circ

A = 9.2^\circ, 99.2^\circ

Question 7

(a)Answer:

R = 13

[2]

R\cos\alpha = 12

R\sin\alpha = 5

R = \sqrt{12^{2} + 5^{2}}

= 13

(b)Answer:

\alpha = 22.62^\circ

[2]

\tan\alpha = \dfrac{5}{12}

\alpha = 22.62^\circ

(c)Answer:

\theta = 67.4^\circ

[2]

Greatest when

\sin(\theta + 22.62^\circ) = 1

\theta + 22.62^\circ = 90^\circ

\theta = 67.4^\circ

(d)Answer:

\theta = 161.8^\circ

\theta = 333.0^\circ

[3]

13\sin(\theta + 22.62^\circ) = -1

\sin(\theta + 22.62^\circ) = \dfrac{-1}{13}

\theta = 161.8^\circ

\theta = 333.0^\circ

Practice questions by skill

Sec 4 A-Math Trigonometric identities and formulae worksheet with answers

In the print window, choose Save as PDF to download it.

Section B

(a)Show that the equation

2\sec^{2} x - \tan x - 3 = 0

can be written as

2\tan^{2} x - \tan x - 1 = 0

[2]

(b)Hence solve the equation

2\sec^{2} x - \tan x - 3 = 0

for

0^\circ \leq x \leq 360^\circ

. Give your answers correct to 1 decimal place.

Answer (b)[4]

(c)Hence state the number of solutions of the equation

2\sec^{2} 2x - \tan 2x - 3 = 0

for

0^\circ \leq x \leq 360^\circ

Answer (c)[1]

Answers and worked solutions

Question 1

Answer:

= \sqrt{34}

[2]

R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha

R\cos\alpha = 3

R\sin\alpha = 5

R = \sqrt{3^{2} + 5^{2}}

= \sqrt{34}

Question 2

(a)Answer:

\cos A = \dfrac{3}{5}

[2]

\text{Adjacent} = \sqrt{5^{2} - 4^{2}}

= 3

A

is in quadrant 1, so

\cos A

is positive.

\cos A = \dfrac{3}{5}

(b)Answer:

\sec A = -\dfrac{5}{4}

[2]

\text{Adjacent} = \sqrt{5^{2} - 3^{2}}

= 4

A

is in quadrant 3, so

\sec A

is negative.

\sec A = -\dfrac{5}{4}

Question 3

(a)Answer:

= \sqrt{13}

[2]

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

R\cos\alpha = 2

R\sin\alpha = 3

R = \sqrt{2^{2} + 3^{2}}

= \sqrt{13}

(b)Answer:

\theta = 295.3^\circ

[3]

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

R\cos\alpha = 3

R\sin\alpha = 4

R = \sqrt{3^{2} + 4^{2}}

= 5

\tan\alpha = \dfrac{4}{3}

\alpha = 53.13^\circ

(2 d.p.)

3\sin\theta + 4\cos\theta = 5\sin(\theta + 53.13^\circ)

\sin(\theta + 53.13^\circ) = -\dfrac{1}{5}

= -0.2000\ldots

\beta = \sin^{-1}\left(0.2000\ldots\right)

\beta = 11.5^\circ

\sin u

is negative in quadrants 3 and 4.

u = 180^\circ + 11.5^\circ

= 191.5^\circ

u = 360^\circ - 11.5^\circ

= 348.5^\circ

\theta = 138.4^\circ

\theta = 295.3^\circ

Question 4

(a)Answer:

= 318.6^\circ

[3]

2\sin x\cos x - 1.5\sin x = 0

\sin x\Big(2\cos x - 1.5\Big) = 0

\sin x = 0

x = 0^\circ

x = 180^\circ

x = 360^\circ

\cos x = \dfrac{3}{4}

\beta = \cos^{-1}\left(0.7500\ldots\right)

\beta = 41.4^\circ

\cos x

is positive in quadrants 1 and 4.

x = 41.4^\circ

x = 360^\circ - 41.4^\circ

= 318.6^\circ

(b)Answer:

= 287.5^\circ

[3]

2\sin x\cos x - 0.6\sin x = 0

\sin x\Big(2\cos x - 0.6\Big) = 0

\sin x = 0

x = 0^\circ

x = 180^\circ

x = 360^\circ

\cos x = \dfrac{3}{10}

\beta = \cos^{-1}\left(0.3000\ldots\right)

\beta = 72.5^\circ

\cos x

is positive in quadrants 1 and 4.

x = 72.5^\circ

x = 360^\circ - 72.5^\circ

= 287.5^\circ

Question 5

(a)[2]

Using

\sec^{2} x = 1 + \tan^{2} x

2(1 + \tan^{2} x) - \tan x - 3 = 0

2\tan^{2} x - \tan x - 1 = 0

(shown)

(b)Answer:

x = 45.0^\circ, 153.4^\circ, 225.0^\circ, 333.4^\circ

[4]

2\tan^{2} x - \tan x - 1 = 0

\tan x = -\dfrac{1}{2}

\tan x = 1

x = 45.0^\circ, 153.4^\circ, 225.0^\circ, 333.4^\circ

(c)Answer:

8

[1]

Each solution for

2x

0^\circ \leq 2x \leq 720^\circ

gives one for

x

.There are

8

solutions.

Question 6

(a)[3]

\cos 2A\cos B + \sin 2A\sin B = 2(\cos 2A\cos B - \sin 2A\sin B)

\sin 2A\sin B + 2\sin 2A\sin B = 2\cos 2A\cos B - \cos 2A\cos B

(3)\sin 2A\sin B = (1)\cos 2A\cos B

\dfrac{\cos 2A\cos B}{\sin 2A\sin B} = 3

\cot 2A\cot B = 3

(shown)

(b)Answer:

\tan 2A = \dfrac{1}{3}

[1]

\cot 45^\circ = 1

, so

\cot 2A = 3

\tan 2A = \dfrac{1}{3}

(c)Answer:

A = 9.2^\circ, 99.2^\circ

[3]

\tan 2A = \dfrac{1}{3}

0^\circ < 2A < 360^\circ

Basic angle

= 18.43^\circ

2A = 18.43^\circ, 198.43^\circ

A = 9.2^\circ, 99.2^\circ

Question 7

(a)Answer:

R = 13

[2]

R\cos\alpha = 12

R\sin\alpha = 5

R = \sqrt{12^{2} + 5^{2}}

= 13

(b)Answer:

\alpha = 22.62^\circ

[2]

\tan\alpha = \dfrac{5}{12}

\alpha = 22.62^\circ

(c)Answer:

\theta = 67.4^\circ

[2]

Greatest when

\sin(\theta + 22.62^\circ) = 1

\theta + 22.62^\circ = 90^\circ

\theta = 67.4^\circ

(d)Answer:

\theta = 161.8^\circ

\theta = 333.0^\circ

[3]

13\sin(\theta + 22.62^\circ) = -1

\sin(\theta + 22.62^\circ) = \dfrac{-1}{13}

\theta = 161.8^\circ

\theta = 333.0^\circ