ABSTRACT
Since 4sin(x − 20◦) = 5cos x then 3.7588sinx − 1.3680cosx = 5cosx Rearranging gives:
3.7588sinx = 5cosx + 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 72 Further problems on compound angle formulae
1. Reduce the following to the sine of one angle:
(a) sin37◦ cos21◦ + cos37◦ sin21◦ (b) sin7t cos3t − cos 7t sin3t
[(a) sin58◦ (b) sin4t ]
2. Reduce the following to the cosine of one angle:
(a) cos71◦ cos33◦ − sin71◦ sin33◦
(b) cos π 3
cos π
4 + sin π
3 sin
π
4⎡ ⎣ (a) cos104◦ ≡ −cos76◦
(b)cos π 12
⎤ ⎦
3. (a) sin
( x + π
+ sin
( x + 2π
= √3cosx
and
(b) −sin (
3π 2
−φ )
= cosφ
4. Prove that: (a) sin
( θ + π
) − sin
( θ − 3π
) =√2(sinθ + cosθ)
(b) cos(270 ◦ +θ)
cos(360◦ −θ) = tanθ
5. Given cos A=0.42 and sin B=0.73 evaluate (a) sin(A− B), (b) cos(A− B), (c) tan(A+B), correct to 4 decimal places.
