Question

Show tat sin 78° + cos 132° = sin 18°.

Answer 1

Before solving the problem, let us know some trigonometric formulae:

• sinC + sinD = 2 * sin{(C + D)/2} * cos{(C + D)/2}

• sinC - sinD = 2 * cos{(C + D)/2} * sin{(C - D)/2}

• cos60° = 1/2

• cos(90° + A) = - sinA

Now we try to prove the given problem:

L.H.S. = sin78° + cos132°

= sin78° + cos(90° + 42°)

= sin78° - sin42°

= 2 * cos{(78° + 42°)/2} * sin{(78° - 42°)/2}

= 2 * cos(120°/2) * sin(36°/2)

= 2 * cos60° * sin18°

= 2 * 1/2 * sin18°

= sin18° = R.H.S.

∴ sin78° + cos132° = sin18°

Hence proved.