ABSTRACT
Problem 5. Solve the equation: 4 sin(x−20◦)= 5cos x for values of x between 0◦ and 90◦
4sin(x− 20◦) = 4[sinxcos20◦ − cos xsin20◦] from the formula forsin(A−B)
= 4[sinx(0.9397)− cos x(0.3420)] = 3.7588sinx− 1.3680cosx
Since 4 sin(x−20◦)=5cosx then 3.7588 sin x−1.3680 cos x=5 cos x Rearranging gives:
3.7588sinx = 5cos x + 1.3680cosx = 6.3680cosx
and sin x cos x
= 6.3680 3.7588
=1.6942
i.e. tan x=1.6942, and x= tan−1 1.6942=59.449◦ or 59◦27′
[Check:LHS = 4sin(59.449◦−20◦) = 4sin39.449◦ = 2.542
RHS = 5cosx = 5cos59.449◦ = 2.542]
Now try the following exercise
Exercise 107 Further problems on compound angle formulae
1. Reduce the following to the sine of one angle: (a) sin37◦ cos21◦+ cos37◦ sin21◦ (b) sin7t cos3t− cos7t sin3t
[(a) sin58◦ (b) sin4t ] 2. Reduce the following to the cosine of one
angle: (a) cos71◦ cos33◦− sin71◦ sin33◦
(b) cos 3
cos
4 + sin
3 sin
4⎡ ⎣(a) cos104◦ ≡ −cos76◦
(b) cos π
⎤ ⎦
3. Show that: (a) sin
( x+ π
) + sin
( x+ 2π
) = √3cos x
(b) −sin (
3π 2
−φ )
= cosφ
4. Prove that:
(a) sin ( θ + π
) − sin
( θ − 3π
) =
√ 2(sin θ + cosθ)
(b) cos(270 ◦ +θ)
cos(360◦ −θ) = tanθ
5. Given cos A=0.42 and sinB=0.73 evaluate (a) sin(A− B), (b) cos(A− B), (c) tan(A+ B), correct to 4 decimal places.
