(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.?

Question

(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

Impact of this question

5904 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-16T10:06:19

$see explanation$ $"see explanation"$

Explanation:

$(a)$ $(a)$

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

$∙ x 2 sin A sin B = cos (A - B) - cos (A + B)$ $•color(white)(x)2sinAsinB=cos(A-B)-cos(A+B)$

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

$sin 7 x sin 3 x$ $sin7xsin3x$

$= \frac{1}{2} (cos (7 x - 3 x) - cos (7 x + 3 x))$ $=1/2(cos(7x-3x)-cos(7x+3x))$

$= \frac{1}{2} (cos 4 x - cos 10 x)$ $=1/2(cos4x-cos10x)$

$(b)$ $(b)$

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

$∙ x tan (x \pm y) = \frac{tan x \pm tan y}{1 \mp tan x tan y}$ $•color(white)(x)tan(x+-y)=(tanx+-tany)/(1∓tanxtany)$

$note that tan (\frac{π}{4}) = 1$ $"note that "tan(pi/4)=1$

$tan (θ + \frac{π}{4}) tan (θ - \frac{π}{4})$ $tan(theta+pi/4)tan(theta-pi/4)$

$= \frac{tan θ + tan (\frac{π}{4})}{1 - tan θ tan (\frac{π}{4})} \times \frac{tan θ - tan (\frac{π}{4})}{1 + tan θ tan (\frac{π}{4})}$ $=(tantheta+tan(pi/4))/(1-tanthetatan(pi/4))xx(tantheta-tan(pi/4))/(1+tanthetatan(pi/4))$

$= \frac{tan θ + 1}{1 - tan θ} \times \frac{tan θ - 1}{1 + tan θ}$ $=(tantheta+1)/(1-tantheta)xx(tantheta-1)/(1+tantheta)$

$= \frac{{tan}^{2} θ - 1}{1 - {tan}^{2} θ}$ $=(tan^2theta-1)/(1-tan^2theta)$

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

Science

Math

Humanities

... and beyond

(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

Explanation:

Related questions

Impact of this question