(a) how to write the product sin(7x) sin(3x) as a sum? (b)how to verify that tan(θ + π/4) tan(θ − π/4) = −1.?

1 Answer
Jul 16, 2018

"see explanation"see explanation

Explanation:

(a)(a)

"using the "color(blue)"product to sum/difference formula"using the product to sum/difference formula

•color(white)(x)2sinAsinB=cos(A-B)-cos(A+B)x2sinAsinB=cos(AB)cos(A+B)

"here "A=7x" and "B=3xhere A=7x and B=3x

sin7xsin3xsin7xsin3x

=1/2(cos(7x-3x)-cos(7x+3x))=12(cos(7x3x)cos(7x+3x))

=1/2(cos4x-cos10x)=12(cos4xcos10x)

(b)(b)

"using the "color(blue)"sum/difference identities for tan"using the sum/difference identities for tan

•color(white)(x)tan(x+-y)=(tanx+-tany)/(1∓tanxtany)xtan(x±y)=tanx±tany1tanxtany

"note that "tan(pi/4)=1note that tan(π4)=1

tan(theta+pi/4)tan(theta-pi/4)tan(θ+π4)tan(θπ4)

=(tantheta+tan(pi/4))/(1-tanthetatan(pi/4))xx(tantheta-tan(pi/4))/(1+tanthetatan(pi/4))=tanθ+tan(π4)1tanθtan(π4)×tanθtan(π4)1+tanθtan(π4)

=(tantheta+1)/(1-tantheta)xx(tantheta-1)/(1+tantheta)=tanθ+11tanθ×tanθ11+tanθ

=(tan^2theta-1)/(1-tan^2theta)=tan2θ11tan2θ

=(-cancel((1-tan^2theta)))/cancel((1-tan^2theta))=-1" as required"