Trigonometric identities for SQA National 5 Maths

Trigonometric identities

This page covers the following topics:

1. Trigonometric identities
2. Trigonometric equations

Two commonly used trigonometric identities are tanx = sinx/cosx and sin²x + cos²x = 1. They can be used to manipulate and simplify trigonometric expressions. The ratio of sine and cosine identity is useful when the ratio of two functions is obtained or when a tangent function is worth expanding. The sum of sine squared and cosine squared can be applied to simplify an expression containing both of the squares.

Trigonometric identities

Trigonometric identities can be used to solve equations involving trigonometric functions. When both sine and cosine functions of the same angle can be separated to be on different sides of an equation, the ratio of the functions can be used to get a tangent to proceed with obtaining values for angles. If both sine and cosine of the same angle are involved in an equation, one of which being squared, the sum of their squares formula can be used to reduce the number of different functions in an equation by using substitution.

Trigonometric equations

1

Use trigonometric identities to give the following expression in terms of cosx: (sinx) ÷ (tanx) + sin²x.

The identities tanx = sinx/cosx and sin²x + cos²x = 1 are used.
(sinx) ÷ (tanx) + sin²x =
= cosxsinx ÷ sinx + sin²x =
= cosx + sin²x =
= cosx + 1 − cos²x

cosx + 1 − cos²x

Use trigonometric identities to give the following expression in terms of cosx: (sinx) ÷ (tanx) + sin²x.

2

Ekin wants to find the solutions to tanxsinx = 3cosx for 0 < x < 180°. Find these solutions.

The identities tanx = sinx/cosx and sin²x + cos²x = 1 must be used.
tanxsinx = 3cosx
sinxsinx ÷ cosx = 3cosx
sin²x = 3cos²x
1 − cos²x = 3cos²x
1 = 4cos²x
cos²x = 1/4
cosx = ±1/2
x = arccos(1/2) or x = arccos(−1/2)
x = 60° or x = 120°

x = 60° or x = 120°

Ekin wants to find the solutions to tanxsinx = 3cosx for 0 < x < 180°. Find these solutions.

3

Give the following expression in terms of tanx and factorise the result: (sinx) ÷ (cosx) − 2tan²x.

The identities tanx = sinx/cosx and sin²x + cos²x = 1 are used.
(sinx) ÷ (cosx) − 2tan²x =
= tanx − 2tan²x =
= tanx(1 − 2tanx)

tanx(1 − 2tanx)

Give the following expression in terms of tanx and factorise the result: (sinx) ÷ (cosx) − 2tan²x.

4

Sarah believes the following equation has one solution: 3sinx = 1/tanx for 0 < x < 180°. Find this solution in degrees.

The identities tanx = sinx/cosx and sin²x + cos²x = 1 must be used.
3sinx = cosx ÷ sinx
3sin²x = cosx
3(1 − cos²x) = cosx
3 − 3cos²x = cosx
3cos²x + cosx − 3 = 0
Using the quadratic equation, cosx = (−1 ± √37) ÷ 6
x = arccos((−1 + √37) ÷ 6) = 32.1°
cosx = (−1 − √37) ÷ 6 has no solutions.

32.1°

Sarah believes the following equation has one solution: 3sinx = 1/tanx for 0 < x < 180°. Find this solution in degrees.

5

Find the solution, in degrees, of the following equation: 3sin²x + 8cosx + 1 = 0 for 0 < x < 180°.

The identity sin²x + cos²x = 1 is used.
3sin²x + 8cosx + 1 = 0
3(1 − cos²x) + 8 cosx + 1 = 0
3 − 3cos²x + 8cosx + 1 = 0
3cos²x − 8cosx − 4 = 0
Using the quadratic equation, cosx = (4 ± 2√7) ÷ 3
cosx = (4 + 2√7) ÷ 3 has no solutions.
x = arccos((4 − 2√7) ÷ 3) = 115° (to 3 s. f.)

115°

Find the solution, in degrees, of the following equation: 3sin²x + 8cosx + 1 = 0 for 0 < x < 180°.

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