∣∣sinθ sin(θ+32π ) sin(θ−32π ) cosθ cos(θ+32π ) cos(θ−32π ) tanθ sin(2θ+34π ) sin(2θ−34π ) ∣∣ equals

Question

∣∣​sinθ  sin(θ+32π ​ )  sin(θ−32π ​ )  ​cosθ  cos(θ+32π ​ )  cos(θ−32π ​ )  ​tanθ  sin(2θ+34π ​ )  sin(2θ−34π ​ )  ​∣∣​ equals

Filo · Accepted Answer

∣∣​sinθsin(θ+32π​) sin(θ−32π​)​cosθ cos(θ+32π​) cos(θ−32π​) ​tanθsin(2θ+34π​)sin(2θ−34π​)​∣∣​R2​→R2​+R3​=∣∣​sinθsin(θ+32π​)+sin(θ−32π​) sin(θ−32π​)​cosθ cos(θ+32π​)+cos(θ−32π​) cos(θ−32π​) ​tanθsin(2θ+34π​)+sin(2θ−34π​)sin(2θ−34π​)​∣∣​We know that sinC+sinD=2sin(2C+D​)cos(2C−D​)and cosC+cosD=2cos(2C+D​)cos(2C−D​)=∣∣​sinθ2sinθcos32π​ sin(θ−32π​)​cosθ 2cosθcos32π​ cos(θ−32π​) ​tanθ2sin2θcos34π​sin(2θ−34π​)​∣∣​We know that sin(−θ)=−sinθ and cos(−θ)=cosθ =∣∣​sinθ2sinθcos32π​ −sin(32π​−θ)​cosθ 2cosθcos32π​ cos(32π​−θ) ​tanθ2sin2θcos34π​−sin(34π​−2θ)​∣∣​=∣∣​sinθ2sinθcos(π−3π​) −sin(π−3π​−θ)​cosθ 2cosθcos(π−3π​) cos(π−3π​−θ) ​tanθ2sin2θcos(π+3π​)−sin(π+3π​−2θ)​∣∣​We know that sin(π−θ)=sinθ,cos(π−θ)=−cosθ,sin(π+θ)=−sinθ,cos(π+θ)=cosθ=∣∣​sinθ−2sinθcos3π​ −sin(π−(3π​+θ))​cosθ −2cosθcos3π​ cos(π−(3π​+θ)) ​tanθ2sin2θcos3π​−sin(π+(3π​−2θ))​∣∣​=∣∣​sinθ−2sinθcos3π​ −sin(3π​+θ)​cosθ −2cosθcos3π​ −cos(3π​+θ) ​tanθ2sin2θcos3π​sin(3π​−2θ)​∣∣​We know that cos3π​=21​=∣∣​sinθ−2sinθ×21​ −sin(3π​+θ)​cosθ −2cosθ×21​ −cos(3π​+θ) ​tanθ2sin2θ×21​sin(3π​−2θ)​∣∣​=∣∣​sinθ−sinθ −sin(3π​+θ)​cosθ −cosθ −cos(3π​+θ) ​tanθsin2θsin(3π​−2θ)​∣∣​R1​→R1​+R2​=∣∣​0−sinθ −sin(3π​+θ)​0 −cosθ −cos(3π​+θ) ​tanθ+sin2θsin2θsin(3π​−2θ)​∣∣​=0−0+(tanθ+sin2θ)(sinθcos(3π​+θ)−cosθsin(3π​+θ))We know that sin(A−B)=sinAcosB−cosAsinB=(tanθ+sin2θ)sin(θ−3π​−θ)=−(tanθ+sin2θ)sin3π​=−23​​(tanθ+sin2θ)

Question Text	$∣ ∣ sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) tan θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) ∣ ∣$ equals
Answer Type	Text solution:1
Upvotes	150

$∣ ∣ sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) tan θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) ∣ ∣$ equals

Text solutionVerified

Practice questions from similar books

∣∣​sinθ sin(θ+32π ​ ) sin(θ−32π ​ ) ​cosθ cos(θ+32π ​ ) cos(θ−32π ​ ) ​tanθ sin(2θ+34π ​ ) sin(2θ−34π ​ ) ​∣∣​ equals

Text solutionVerified

Practice questions from similar books

$∣ ∣ sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) tan θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) ∣ ∣$ equals