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sweet-ann [11.9K]
2 years ago
9

Verify (2cos2x)/(sin2x) =cotx - tanx

Mathematics
1 answer:
lianna [129]2 years ago
6 0
Cos2x = (cosx)^2 - (sinx)^2;
sin2x = 2sinxcosx;
Then, (2cos2x)/(sin2x) =2[ (cosx)^2 - (sinx)^2 ] / (2sinxcosx) = [ (cosx)^2 - (sinx)^2 ] / (sinxcosx) = (cosx)^2 / (sinxcosx) - (sinx)^2 / (sinxcosx) = cosx/sinx - sinx/cosx = cotx - tanx;
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lutik1710 [3]

Answer:

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Step-by-step explanation:

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We use the law of sines in the blue triangle to do such:

\frac{sin(z)}{11} =\frac{sin(133)}{20} \\sin(z)=\frac{11\,sin(133)}{20} \\sin(z)=0.4022

Now we can use this value in the larger right angle triangle where WX is the opposite side to angle  \angle z, and the 20 mm side is the hypotenuse:

sin(z)=\frac{opposite}{hypotenuse} \\sin(z)=\frac{WX}{20}\\0.4022=\frac{WX}{20}\\WX=20\,(0.4022)\\WX=8.044\,\,mm

which rounded to the nearest integer gives

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son4ous [18]

Answer:

x = 4

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