Question

find the exact value using a sum or difference identity. cos 240° hint: sin(a ± b) = sin a cos b ± cos a sin b cos(a ± b) = cos a cos b ∓ sin a sin b

Explanation:

Step1: Express \(240^{\circ}\) as a sum

Express \(240^{\circ}\) as \(180^{\circ}+60^{\circ}\).
So, \(\cos240^{\circ}=\cos(180^{\circ} + 60^{\circ})\)

Step2: Apply the cosine sum identity

The cosine sum identity is \(\cos(A + B)=\cos A\cos B-\sin A\sin B\).
Here \(A = 180^{\circ}\), \(B=60^{\circ}\).
\(\cos(180^{\circ}+60^{\circ})=\cos180^{\circ}\cos60^{\circ}-\sin180^{\circ}\sin60^{\circ}\)
We know that \(\cos180^{\circ}=- 1\), \(\cos60^{\circ}=\frac{1}{2}\), \(\sin180^{\circ}=0\), \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\)
Substitute these values:

$$
LATEXBLOCK0
$$

Answer:

\(-\frac{1}{2}\)