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sinA+Sin2A+sin3A = sinBcosA+cos2A+cos3A = cosB then , expresS A in terms of B

PRANAV SATHEESH , 9 Years ago
Grade 11
anser 2 Answers
Lab Bhattacharjee

Last Activity: 9 Years ago

HINT:
 $$\sin A+\sin3A=2\sin\dfrac{A+3A}2\cos\dfrac{3A-A}2 \\ \text{ and }\cos A+\cos3A=2\cos\dfrac{A+3A}2\cos\dfrac{3A-A}2$$

rahul

Last Activity: 7 Years ago

SinA+sin3A+sin2A=sinB2sin(A+3A)/2*cos(3A-A)+sin2A=sinB2sin2A*cosA+sin2A=sinB-----(1)Similarly in case of cos2cos2A*cosA+cos2A=cosB------(2)Dividing 1&2Sin2A(cosA+1)/cos2A(cosA+1)=sinB/cosBTan2A=tan B2A=BA=B/2

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