# Trigonometry Questions

## If A + B + C = $$3\pi\over 2$$, then cos2A + cos2B + cos2C is equal to

Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C = 2cos($$3\pi\over 2$$ – C)cos(A-B) + cos2C  $$\because$$  A + B + C = $$3\pi\over 2$$ = -2sinC cos(A-B) + 1 – 2$$sin^2C$$ = 1 – 2sinC[cos(A-B)+sinC] = 1 – 2sinC[cos(A-B) + sin($$3\pi\over 2$$-(A+B))] = 1 – 2sinC[cos(A-B)-cos(A+B)] = 1 – 4sinA sinB sinC …

## $$sin5x + sin2x – sinx\over {cos5x + 2cos3x + 2cos^x + cosx}$$ is equal to

Solution : L.H.S. = $$2sin2xcos3x + sin2x\over{2cos3x.cos2x + 2cos3x + 2cos^2x}$$ = $$sin2x[2cos3x+1]\over {2[cos3x(cos2x+1)+(cos^2x)]}$$ = $$sin2x[2cos3x+1]\over {2[cos3x(2cos^2x)+(cos^2x)]}$$ = $$sin2x[2cos3x+1]\over {2cos^2x(2cos3x+1)}$$ = tanx Similar Questions Evaluate sin78 – sin66 – sin42 + sin6. If A + B + C = $$3\pi\over 2$$, then cos2A + cos2B + cos2C is equal to Find the maximum value of …

## Prove that $$2cos2A+1\over {2cos2A-1}$$ = tan($$60^{\circ}$$ + A)tan($$60^{\circ}$$ – A)

Solution : R.H.S. = tan($$60^{\circ}$$ + A)tan($$60^{\circ}$$ – A) = ($$tan60^{\circ}+tanA\over {1-tan60^{\circ}tanA}$$)($$tan60^{\circ}-tanA\over {1+tan60^{\circ}tanA}$$) = ($$\sqrt{3}+tanA\over {1-\sqrt{3}tanA}$$)($$\sqrt{3}-tanA\over {1+\sqrt{3}tanA}$$) = $$3-tan^2A\over{1-3tan^2A}$$ = $$3cos^2A-sin^2A\over {cos^2A-3sin^2A}$$ = $$2cos^2A+cos^2A-2sin^2A+sin^2A\over {2cos^2A-2sin^2A-sin^2A-cos^2A}$$ = $$2(cos^2A-sin^2A)+cos^2A+sin^2A\over {2(cos^2A-sin^2A)-(sin^2A+cos^2A)}$$ = $$2cos2A+1\over {2cos2A-1}$$ = L.H.S Similar Questions Evaluate sin78 – sin66 – sin42 + sin6. If A + B + C = $$3\pi\over 2$$, then cos2A + …