Question

With usual notations, in a triangle ABC, if θ is any real number, then a cos(B - θ) + b cos (A + θ) is 

Options

• a cosθ

• b cosθ

• cosθ

• c cosθ

Answer 1

c cosθ

Explanation:

a cos (B − θ) + b cos (A + θ)

= a (cos B cos θ + sin B sin θ) + b (cos A cos θ − sin A sin θ)

= a cos B cos θ + a sin B sin θ + b cos A cos θ − b sin A sin θ

= cos θ (a cos B + b cos A) + sin θ (a sin B − b sin A)

\[=\cos\theta\left[\frac{\mathbf{a}\left(\mathbf{a}^2+\mathbf{c}^2-\mathbf{b}^2\right)}{2\mathbf{a}\mathbf{c}}+\frac{\mathbf{b}\left(\mathbf{b}^2+\mathbf{c}^2-\mathbf{a}^2\right)}{2\mathbf{b}\mathbf{c}}\right]\] + sin θ (abk − abk) ...[by cosine and since rule]

\[=\cos\theta\left(\frac{\mathbf{a}^{2}+\mathbf{c}^{2}-\mathbf{b}^{2}+\mathbf{b}^{2}+\mathbf{c}^{2}-\mathbf{a}^{2}}{2\mathbf{c}}\right)\]

c cosθ