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Trigonometry Questions

Find the maximum value of 1 + sin(π4+θ) + 2cos(π4θ)

Solution : We have 1 + sin(π4+θ) + 2cos(π4θ) = 1 + 1sqrt2(cosθ+cosθ) + 2(cosθ+cosθ) = 1 + (12+2) + (cosθ+cosθ) = 1 + (12+2).2cos(θpi4)    Maximum Value = 1 + \(({1\over \sqrt{2}}

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Evaluate sin78 – sin66 – sin42 + sin6.

Solution : The expression = (sin78 – sin42) – (sin66 – sin6) = 2cos(60)sin(18) – 2cos36.sin30 = sin18 – cos36 = (514)(5+14) = -12 Similar Questions Find the maximum value of 1 + sin(π4+θ) + 2cos(π4θ) If A + B

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If A + B + C = 3π2, then cos2A + cos2B + cos2C is equal to

Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C = 2cos(3π2 – C)cos(A-B) + cos2C    A + B + C = 3π2 = -2sinC cos(A-B) + 1 – 2sin2C = 1 – 2sinC[cos(A-B)+sinC] = 1 – 2sinC[cos(A-B) + sin(3π2-(A+B))] = 1 – 2sinC[cos(A-B)-cos(A+B)] = 1 – 4sinA sinB sinC

If A + B + C = 3π2, then cos2A + cos2B + cos2C is equal to Read More »