Cos 2theta formula

Here are a couple of methods... Method 1 Using: #cos 2 theta = cos^2 theta - sin^2 theta# #cos^2 theta + sin^2 theta = 1# So: #cos 2 theta = cos^2 theta - sin^2 theta = cos^2 theta - (1 - cos^2 theta) = 2cos^2 theta - 1# So: #cos 4 theta = 2cos^2 2 theta - 1# #color(white)(cos 4 theta) = 2(2cos^2 theta - 1)^2 - 1# #color(white)(cos 4 theta) = 2(4cos^4 theta - 4cos^2 theta + 1) - 1# #color(white)(cos 4 theta) = 8cos^4 theta - 8cos^2 theta + 1# Method 2 Using de Moivre's theorem: #(cos theta + i sin theta)^n = cos n theta + i sin n theta# and #cos^2 theta + sin ^2 theta = 1# So: #cos 4 theta + i sin 4 theta# #= (cos theta + i sin theta)^4# #= cos^4 theta + 4i cos^3 theta sin theta - 6 cos^2 theta sin^2 theta - 4i cos theta sin^3 theta + sin^4 theta# Equating real parts: #cos 4 theta = cos^4 theta - 6 cos^2 theta sin^2 theta + sin^4 theta# #color(white)(cos 4 theta) = cos^4 theta - 6 cos^2 theta (1-cos^2 theta) + (1-cos^2 theta)^2# #color(white)(cos 4 theta) = cos^4 theta - 6 cos^2 theta + 6 cos^4 theta + 1-2cos^2 theta+cos^4 theta# #color(white)(cos 4 theta) = 8cos^4 theta - 8 cos^2 theta + 1#

#cos2theta=costheta# or #2cos^2theta-1=costheta# (using formula for #cos2theta#) the above becomes #2cos^2theta-costheta-1=0# Now using quadratic formula #costheta=(-(-1)+-sqrt((-1)^2-4*2*(-1)))/(2*2)# or #costheta=(1+-sqrt(1+8))/4=(1+-3)/4# Hence #costheta=1=cos0# or #costheta=-1/2=cos((2pi)/3)# Hence #theta=(2n+1)pi/2# or #theta=2npi+-(2pi)/3#, where #n# is an integer.

The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. Triple-angle Identities \[ \sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta \] \[ \cos 3\theta = 4 \cos ^ 3 \theta - 3 \cos \theta \] To prove the triple-angle identities, we can write \(\sin 3 \theta\) as \(\sin(2 \theta + \theta)\). Then we can use the \[\begin \] Prove that \[ \tan (6^\circ) = \tan (12^\circ) \tan(24^\circ) \tan(48^\circ ). \]

Formula Summary We derive the following formulas on this page: `sin (alpha/2)=+-sqrt((1-cos alpha)/2` `cos (alpha/2)=+-sqrt((1+cos alpha)/2` `tan (alpha/2)=(1-cos alpha)/(sin alpha` Now, if we let `theta=alpha/2` then 2 θ = α and our formula becomes: `cos α = 1 − 2\ sin^2(α/2)` We now solve for `sin(alpha/2)` (That is, we get `sin(alpha/2)` on the left of the equation and everything else on the right): `2\ sin^2(α/2) = 1 − cos α` `sin^2(α/2) = (1 − cos α)/2` Solving gives us the following sine of a half-angle identity: `sin (alpha/2)=+-sqrt((1-cos alpha)/2` The sign (positive or negative) of `sin(alpha/2)` depends on the quadrant in which `α/2` lies. If `α/2` is in the first or second quadrants, the formula uses the positive case: `sin (alpha/2)=sqrt(1-cos alpha)/2` If `α/2` is in the third or fourth quadrants, the formula uses the negative case: `sin (alpha/2)=-sqrt(1-cos alpha)/2` Half Angle Formula - Cosine Using a similar process, with the same substitution of `theta=alpha/2` (so 2 θ = α) we subsitute into the identity cos 2 θ = 2cos 2 θ− 1 (see We obtain `cos alpha=2\ cos^2(alpha/2)-1` Reverse the equation: `2\ cos^2(alpha/2)-1=cos alpha` Add 1 to both sides: `2\ cos^2(alpha/2)=1+cos alpha` Divide both sides by `2` `cos^2(alpha/2)=(1+cos alpha)/2` Solving for `cos(α/2)`, we obtain: `cos (alpha/2)=+-sqrt((1+cos alpha)/2` As before, the sign we need depends on the quadrant. If `α/2` is in the first or fourth quadrants, the formula uses the positive case: `cos (alpha/2)=...

The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. Triple-angle Identities \[\begin \]

• • • • Formula $\cos^2$ In this way, you can write the cosine squared power reducing trigonometric identity in terms of any symbol. Proof Learn how to prove the cosine squared power reduction trigonometric identity in trigonometry.

The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. Triple-angle Identities \[\begin \]

The trigonometric triple-angle identities give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself. Triple-angle Identities \[ \sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta \] \[ \cos 3\theta = 4 \cos ^ 3 \theta - 3 \cos \theta \] To prove the triple-angle identities, we can write \(\sin 3 \theta\) as \(\sin(2 \theta + \theta)\). Then we can use the \[\begin \] Prove that \[ \tan (6^\circ) = \tan (12^\circ) \tan(24^\circ) \tan(48^\circ ). \]

#cos2theta=costheta# or #2cos^2theta-1=costheta# (using formula for #cos2theta#) the above becomes #2cos^2theta-costheta-1=0# Now using quadratic formula #costheta=(-(-1)+-sqrt((-1)^2-4*2*(-1)))/(2*2)# or #costheta=(1+-sqrt(1+8))/4=(1+-3)/4# Hence #costheta=1=cos0# or #costheta=-1/2=cos((2pi)/3)# Hence #theta=(2n+1)pi/2# or #theta=2npi+-(2pi)/3#, where #n# is an integer.

Formula Summary We derive the following formulas on this page: `sin (alpha/2)=+-sqrt((1-cos alpha)/2` `cos (alpha/2)=+-sqrt((1+cos alpha)/2` `tan (alpha/2)=(1-cos alpha)/(sin alpha` Now, if we let `theta=alpha/2` then 2 θ = α and our formula becomes: `cos α = 1 − 2\ sin^2(α/2)` We now solve for `sin(alpha/2)` (That is, we get `sin(alpha/2)` on the left of the equation and everything else on the right): `2\ sin^2(α/2) = 1 − cos α` `sin^2(α/2) = (1 − cos α)/2` Solving gives us the following sine of a half-angle identity: `sin (alpha/2)=+-sqrt((1-cos alpha)/2` The sign (positive or negative) of `sin(alpha/2)` depends on the quadrant in which `α/2` lies. If `α/2` is in the first or second quadrants, the formula uses the positive case: `sin (alpha/2)=sqrt(1-cos alpha)/2` If `α/2` is in the third or fourth quadrants, the formula uses the negative case: `sin (alpha/2)=-sqrt(1-cos alpha)/2` Half Angle Formula - Cosine Using a similar process, with the same substitution of `theta=alpha/2` (so 2 θ = α) we subsitute into the identity cos 2 θ = 2cos 2 θ− 1 (see We obtain `cos alpha=2\ cos^2(alpha/2)-1` Reverse the equation: `2\ cos^2(alpha/2)-1=cos alpha` Add 1 to both sides: `2\ cos^2(alpha/2)=1+cos alpha` Divide both sides by `2` `cos^2(alpha/2)=(1+cos alpha)/2` Solving for `cos(α/2)`, we obtain: `cos (alpha/2)=+-sqrt((1+cos alpha)/2` As before, the sign we need depends on the quadrant. If `α/2` is in the first or fourth quadrants, the formula uses the positive case: `cos (alpha/2)=...