# PROVING PROBLEMS INVOLVING TRIGONOMETRIC ANGLES FOR GRADE 11

Proving Problems Involving Trigonometric Angles for Grade 11 :

In this section, you will learn how to do proving problems involving trigonometric angles.

## Questions

Question 1 :

Prove that

[cot (180° + θ) sin (90° - θ) cos (- θ)] / [ sin (270° + θ) cot (180° + θ) tan (- θ) cosec (360° + θ)  =  cosθ cot θ

Solution :

L.H.S

cot (180° + θ)

Lies in 3rd quadrant. For tan and cot we will have positive sign.

cot (180° + θ)   =  -cot θ

sin (90° - θ)

Lies in 1st quadrant. For all trigonometric ratios, we will have positive sign.

sin (90° - θ)  =  cos θ

cos (- θ)

According to the property cos (- θ)  =  cos θ

sin (270° + θ)

Lies in 4th quadrant. For cos and sec we will have positive sign.

sin (270° + θ)  =  cos θ

tan (- θ)

According to the property tan (- θ)  =  -tan θ

cosec (360° + θ)

Lies in 1st quadrant. For all trigonometric ratios, we will have positive sign.

cosec (360° + θ)  =  cosec θ

=   -cot θ cos θ cos θ /cos θ (-tan θ) cosec θ

=  (cot θ cos θ) / (tan θ cosec θ) =   cosθ cot θ  -------> R.H.S

Hence proved.

Question 2 :

Find all the angles between 0° and 360°  which satisfy the equation sin2 θ = 3/4.

Solution :

sin2 θ = 3/4

sin θ = √(3/4)

sin θ = 3/2

 θ = sin-1 √3/2 θ  = π/3 In 2nd quadrant, we will have positive values for the trigonometric ratios sin θ and cosec θrequired angle  =  π - (π/3)  =  2π/3

Hence the required angles are π/3 and 2π/3.

Question 3 :

Show that sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9  =  2.

Solution :

L.H.S

=  sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9

=  (sin π/18)2 + (sin π/9)2 + (sin 7π/18)2 + (sin 4π/9)2

=  sin2 10 + sin2 20 + sin 2 70 + sin2 80

=  [cos(90 - 10)]2 + [cos(90 - 20)] + sin 2 70 + sin2 80

=  cos80 + cos70  + sin 2 70 + sin2 80

=  1 + 1

=  2  ------> R.H.S

Hence proved. After having gone through the stuff given above, we hope that the students would have understood how to do proving problems involving trigonometric angles

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