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 = \(\frac{-1+\sqrt{5}}{4}\).