# Compound Angles – Problems

### Compound Angles – Problems

Definition: The algebraic sum of two or more angles is called a Compound angles. i.e, A + B, A – B, A + B + C, A + B – C, A – B + C, B + C – A … etc,

1. Simplify the following

i. cos (100) cos (40) + sin (100) sin (40)

Solution: Given that,

cos (100) cos (40) + sin (100) sin (40)

we can use the formula is cos (A – B) = cos A cos B + sin A sin B

now given that in the form of cos(A-B)

A = 100 and B = 40

cos (100 – 40) = cos (100) cos (40) + sin (100) sin (40)

cos (60) = ½

ii. tan (π/4 + θ) tan (π/4 – θ)

Solution: Given that,

tan (π/4 + θ) tan (π/4 – θ)

we can use tan (A + B) = (tan A + tan B)/ (1- tan A tan B)

tan (π/4 + θ) = (tan (π/4) + tanθ)/(1- tan (π/4) tanθ)

= (1 + tanθ)/ (1 – tanθ) … (1)

tan (π/4 – θ) = (tan(π/4) – tanθ)/ (1+ tan(π/4) tanθ)

= (1 – tanθ)/ (1 + tanθ) … (2)

From equation (1) and (2)

tan (π/4 + θ) tan (π/4 – θ) = (1 + tanθ)/ (1 – tanθ) (1 – tanθ)/ (1 + tanθ)

tan (π/4 + θ) tan (π/4 – θ) = 1

2. Find the values of sin 18

Solution: Given sin 18

Let θ = 18 then

2θ = 36 = 90 – 3θ.

Now sin2θ = 2sinθ cosθ and

sin (90 – 3θ) = cos3θ = 4cos³θ – 3cosθ

2sinθ cosθ = cosθ (4cos²θ – 3) = cosθ (1 – 4sin²θ).

Hence, 2sinθ = 1 – 4sin²θ

4sin²θ + 2sinθ – 1 = 0

We can compare to quadratic expression ax² + bx + c = 0 then find the factors sinθ

=  $$\frac{-2\pm \sqrt{{{2}^{2}}-4(4)(-1)}}{2.4}$$,

= $$\frac{-2\pm \sqrt{4+16}}{2.4}$$,

= $$\frac{-2\pm \sqrt{20}}{2.4}$$,

= $$\frac{-2\pm 2\sqrt{5}}{2.4}$$,

= $$\frac{-1\pm \sqrt{5}}{4}$$,

sinθ = $$\frac{-1+\sqrt{5}}{4}$$, $$\frac{-1-\sqrt{5}}{4}$$,

But as sinθ > 0 we have sinθ =  $$\frac{-1+\sqrt{5}}{4}$$.