# PROVING PROBLEMS INVOLVING TRIGONOMETRIC ANGLES FOR GRADE 11

Problem 1 :

Prove that

[cot(180° + θ)sin(90° - θ)cos(- θ)]/[sin(270° + θ)cot(180° + θ)tan (- θ)cosec(360° + θ)  =  cos2θcotθ

Solution :

cot(180° + θ) :

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

cot(180° + θ)   =  -cotθ

sin(90° - θ) :

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

sin(90° - θ)  =  cosθ

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)  =  cosθ

tan(- θ) :

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

cosec(360° + θ) :

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

cosec(360° + θ)  =  cosecθ

Then,

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

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

=  (cot θ cos θ) / (tan θ cosec θ)

=   cosθ cot θ

Problem 2 :

Find all the angles between 0° and 360°  which satisfy the equation sin2θ = 3/4.

Solution :

sin2θ  =  3/4

sinθ  =  √(3/4)

sinθ  =  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.

Problem 3 :

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

Solution :

sin2π/18 + sin2π/9 + sin27π/18 + sin24π/9 :

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

=  sin210 + sin220 + sin 270 + sin280

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

=  cos280 + cos270  + sin270 + sin280

=  sin280 + cos280 + sin270 + cos270

=  1 + 1

=  2

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