use identities to find values of the sine and cosine functions of the function for the angle measure.\n2x…

use identities to find values of the sine and cosine functions of the function for the angle measure.\n2x, given tanx = 4 and cosx > 0\ncos2x = \n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\nsin2x = \n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

use identities to find values of the sine and cosine functions of the function for the angle measure.\n2x, given tanx = 4 and cosx > 0\ncos2x = \n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)\nsin2x = \n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Answer

Explanation:

Step1: Find (\sin x) and (\cos x)

Since (\tan x=\frac{\sin x}{\cos x} = 4), so (\sin x = 4\cos x). Using the identity (\sin^{2}x+\cos^{2}x = 1), substitute (\sin x = 4\cos x) into it: ((4\cos x)^{2}+\cos^{2}x=1) (16\cos^{2}x+\cos^{2}x = 1) (17\cos^{2}x=1) (\cos^{2}x=\frac{1}{17}) Because (\cos x>0), so (\cos x=\frac{1}{\sqrt{17}}=\frac{\sqrt{17}}{17}) Then (\sin x = 4\cos x=\frac{4\sqrt{17}}{17})

Step2: Use double - angle formulas

The double - angle formula for (\cos2x) is (\cos2x=\cos^{2}x-\sin^{2}x) Substitute (\sin x=\frac{4\sqrt{17}}{17}) and (\cos x=\frac{\sqrt{17}}{17}) into it: (\cos2x=\left(\frac{\sqrt{17}}{17}\right)^{2}-\left(\frac{4\sqrt{17}}{17}\right)^{2}=\frac{17}{289}-\frac{272}{289}=\frac{1 - 16}{17}=-\frac{15}{17})

The double - angle formula for (\sin2x) is (\sin2x = 2\sin x\cos x) Substitute (\sin x=\frac{4\sqrt{17}}{17}) and (\cos x=\frac{\sqrt{17}}{17}) into it: (\sin2x=2\times\frac{4\sqrt{17}}{17}\times\frac{\sqrt{17}}{17}=\frac{8\times17}{289}=\frac{8}{17})

Answer:

(\cos2x=-\frac{15}{17}) (\sin2x=\frac{8}{17})