# Use of Trigonometric Functions of Standard Angles

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### Evaluate the following

#### Example – 01:

sin20c + sin2(π/6)c + sin2(π/3)c + sin2(π/2)c   = (0)2 + (1/2)2 + (3/2)2 + (1)2

= 0 + 1/4 + 3/4 + 1 = 1+ 1 = 2

#### Example – 02:

cos20 + cos2(π/6)c + cos2(π/3)c + cos2(π/2)c   = (1)2 + (3/2)2 + (1/2)2 + (0)2

= 1 + 3/4 + 1/4 + 0 = 1+ 1 = 2

#### Example – 03:

sin (π)c + 2 cos (π)c + 3 sin (3π/2)c + 4 cos (3π/2)c – 5 sec (π)c + 6 cosec (π/2)c

=  (0) + 2(-1) + 3(-1) + 4 (0) – 5 (-1) + 6 (1) = 0 – 2 – 3 + 5 + 6 =  6

#### Example – 04:

sin (0)c + 2 cos (0)c + 3 sin (π/2)c + 4 cos (π/2)c + 5 sec (0)c + 6 cosec (π/2)c

= 0 + 2 (1) + 3 (1) + 4 (0) + 5 (1) + 6 (1) = 2 + 3 + 0 + 5 + 6 = 16

#### Example – 05:

4 cot 45° – sec2 60° + sin2 30° = 4 (1) – (2)2 + (1/2)2  =  4  – 4 + 1/4  = 1/4

### Verify the following

#### Example – 06:

• cot2 60° + sin2 45° + sin2 30° + cos2 90° = 13/12

L.H.S. = cot2 60° + sin2 45° + sin2 30° + cos2 90°

∴ L.H.S. = (1/3)2  + (1/2)2 + (1/2)2 + (0)2

∴ L.H.S. =  1/3 + 1/2 + 1/4 = 5/6 + 1/4 = 26/24 = 13/12 = R.H.S.

∴   cot2 60° + sin2 45° + sin2 30° + cos2 90° = 13/12  (Proved)

#### Example – 07:

• sin2 30° + cos2 60° + tan2 45° + sec2 60°  – cosec2 30° = 3/2

L.H.S. = sin2 30° + cos2 60° + tan2 45° + sec2 60°  – cosec2 30°

∴ L.H.S. = (1/2)2  + (1/2)2 + (1)2  + (2)2 – (2)2

∴ L.H.S. =  1/4 + 1/4 + 1 + 4 – 4 = 2/4 +1 = 1/2 + 1 = 3/2  = R.H.S.

∴  sin2 30° + cos2 60° + tan2 45° + sec2 60°  – cosec2 30° = 3/2   (Proved)

#### Example – 08:

• 4 cot2 30° + 9 sin2 60° – 6 cosec2 60°  -(9/4) tan2 60° = 4

L.H.S. = 4 cot2 30° + 9 sin2 60° – 6 cosec2 60°  – (9/4) tan2 60°

∴ L.H.S. = 4 (3)2  + 9 (3/2)2 – 6 (2/3)2 – (9/4)(3) 2

∴ L.H.S. = 4 x 3  + 9 (3/4) – 6 (4/3) – (9/4)(3)

∴ L.H.S. = 12  + 27/4 – 8 – 27/4 = 4 = R.H.S,

∴ 4 cot2 30° + 9 sin2 60° – 6 cosec2 60°  -(9/4) tan2 60° = 4 (Proved)

#### Example – 11:

• If 2 cos2θ + 3cosθ = 2, then find cosθ.
• Solution:

Given 2 cos2θ + 3cosθ = 2  i.e 2 cos2θ + 3cosθ – 2 = 0

let cosθ = x

∴ 2 x2 + 3x – 2 = 0

∴ 2 x2 + 4x – x – 2 = 0

∴ 2 x(x + 2) – 1(x + 2) = 0

∴ (x + 2)(2x – 1) = 0

∴ (x + 2) = 0 or (2x – 1) = 0

∴ x = – 2 or x = 1/2

∴ cosθ = – 2 or cosθ = 1/2

Now – 1 ≤ cos θ ≤ 1, thus cosθ = – 2 is not possible.

∴ cosθ = 1/2

#### Example – 12:

• If 6sin2θ  – 11sinθ + 4 = 0, find cosθ.
• Solution:

Given 6sin2θ  – 11sinθ + 4 = 0

let sinθ = x

∴ 6x2  – 11x + 4 = 0

∴ 6x2  – 8x – 3x + 4 = 0

∴ 2x(3x  – 4) – 1(3x – 4) = 0

∴ (3x  – 4)(2x – 1) = 0

∴ 3x  – 4 = 0 or 2x – 1 = 0

∴ x = 4/3 or x = 1/2

Now – 1 ≤ sin θ ≤ 1, thus sinθ = 4/3 is not possible.

∴ sinθ = 1/2

#### Example – 13:

• If 3tan2θ  – 43 tanθ + 3 = 0, then find tanθ.
• Solution:

Given 3tan2θ  – 43 tanθ + 3 = 0

let tanθ = x

∴ 3x2  – 43 x + 3 = 0

∴ 3x2  – 33 x –3 x + 3 = 0

∴ √3 x 3 x2  – 33 x –3 x + 3= 0

∴ √3 x ( 3x – 3 )  – 1(3 x – 3) = 0

∴  (3x – 3 )( 3 x – 1) = 0

∴ 3x – 3  = 0 or 3 x – 1 = 0

∴ x = 3/3 or x = 1/3

∴ x = 3 or x = 1/3

∴  tanθ = 3 or tanθ = 1/3

#### Example – 14:

• If 4sin2θ  – 2(3 + 1)sinθ + 3 = 0, then find sinθ. Hence find the angle θ.
• Solution:

Given 4sin2θ  – 2(3 + 1)sinθ + 3 = 0

let sinθ = x

∴ 4x2  – 2(3 + 1)x + 3 = 0

∴ 4x2  – 23x – 2x + 3 = 0

∴ 2x(2x  – 3) – 1(2x – 3 ) = 0

∴ (2x  – 3)(2x – 1) = 0

∴ 2x  – 3  = 0 or 2x – 1 = 0

∴ x = 3/2 or x = 1/2

∴  sinθ = 3/2 or sinθ = 1/2

∴ θ = sin-1(3/2) or θ = sin-1(1/2)

∴ θ = (π/3)c i.e. 60° or θ = (π/6)c i.e. 30°

#### Example – 15:

• Find the acute angles A and B satisfying cot (A + B) = 1 and cosec (A- B) = 2
• Solution:

Given cot (A + B) = 1 and cosec (A- B) = 2

we know that cot 45° = 1 and cosec 30° = 2

∴ A + B = 45° and A- B = 30°

Adding the two equations we get

2A = 75° i.e. A = 37.5°

Subtracting the two equations we get

2B = 15° i.e. B = 7.5°

∴ A = 37.5° and B = 7.5°

#### Example – 16:

• Find the acute angles secA.cotB – secA – 2 cotB + 2 = 0
• Solution:

Given secA.cotB – secA – 2 cotB + 2 = 0

secA.(cotB – 1) – 2 (cotB – 1) = 0

(cotB – 1)(secA – 2) = 0

∴ cotB – 1 = 0  and/or secA – 2 = 0

∴ cotB = 1 and/or secA = 2

∴B =  cot-11 and/or A = sec-12

∴B = (π/4)c i.e. 45° and/or A = (π/3)c i.e. 60°

∴ A = (π/3)c i.e. 60° and B = (π/4)c i.e. 45°

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