How do you derive the trigonometric formulas for double and half angels for sin, cos, and tan? I.e: How do I derive something like sin(2x)=2(sinx)(cosx)?

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How do you derive the trigonometric formulas for double and half angels for sin, cos, and tan? I.e: How do I derive something like sin(2x)=2(sinx)(cosx)?

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Answer 1 · 2018-06-16T07:17:14

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Explanation:

Remember that $2 x = x + x$ $2x=x+x$

$sin (2 x) = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x$ $sin(2x)=sin(x+x)=sinxcosx+cosxsinx=color (red) (2sinxcosx)$

$cos (2 x) = cos (x + x) = cos x cos x - sin x sin x = {cos}^{2} x - {sin}^{2} x$ $cos(2x)=cos(x+x)=cosxcosx-sinxsinx=color (red) (cos^2x-sin^2x$

Now let $2 x = θ$ $2x=theta$ so $x = \frac{θ}{2}$ $x=theta/2$

$cos θ = {cos}^{2} (\frac{θ}{2}) - {sin}^{2} (\frac{θ}{2})$ $costheta=cos^2(theta/2)-sin^2(theta/2)$

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

$2 {cos}^{2} (\frac{θ}{2}) = cos θ + 1$ $2cos^2(theta/2)=costheta+1$

${cos}^{2} (\frac{θ}{2}) = \frac{1}{2} (cos θ + 1)$ $cos^2(theta/2)=1/2(costheta+1)$

$cos (\frac{θ}{2}) = \pm \sqrt{\frac{1}{2} (cos θ + 1)}$ $color (red) (cos(theta/2)=+-sqrt(1/2(costheta+1)$

Similarly,

$cos θ = 1 - 2 {sin}^{2} (\frac{θ}{2})$ $costheta=1-2sin^2(theta/2)$

$2 {sin}^{2} (\frac{θ}{2}) = 1 - cos θ$ $2sin^2(theta/2)=1-costheta$

${sin}^{2} (\frac{θ}{2}) = \frac{1}{2} (1 - cos θ)$ $sin^2(theta/2)=1/2(1-costheta)$

$sin (\frac{θ}{2}) = \pm \sqrt{\frac{1}{2} (1 - cos θ)}$ $color (red) (sin(theta/2)=+-sqrt(1/2(1-costheta)$

$tan (2 x) = tan (x + x) = \frac{tan x + tan x}{1 - {tan}^{2} x} = \frac{2 tan x}{1 - {tan}^{2} x}$ $tan(2x)=tan(x+x)=(tanx+tanx)/(1-tan^2x)=color (red) ((2tanx)/(1-tan^2x)$

Finally $tan (\frac{θ}{2}) = \frac{sin (\frac{θ}{2})}{cos (\frac{θ}{2})}$ $tan(theta/2)=sin(theta/2)/cos(theta/2)$

$tan (\frac{θ}{2}) = \frac{\pm \sqrt{\frac{1}{2} (1 - cos θ)}}{\pm \sqrt{\frac{1}{2} (cos θ + 1)}}$ $tan(theta/2)=(+-sqrt(1/2(1-costheta)))/(+-sqrt(1/2(costheta+1))$

$tan (\frac{θ}{2}) = \frac{\sqrt{1 - cos θ}}{\sqrt{1 + cos θ}}$ $tan(theta/2)=sqrt(1-costheta)/sqrt(1+costheta)$

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

$tan (\frac{θ}{2}) = \frac{sin θ}{1 + cos θ}$ $color (red) (tan(theta/2)=sintheta/(1+costheta)$

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How do you derive the trigonometric formulas for double and half angels for sin, cos, and tan? I.e: How do I derive something like sin(2x)=2(sinx)(cosx)?

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