Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0
+26400 Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0
In trigonometry, the tangent half-angle formulas relate the tangent of one half of an angle to trigonometric functions of the entire angle. They are as follows:
\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} \sin{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } } \\\\ \cos{(\alpha)} &=& \dfrac{ 1 - \tan^2{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } }\\\\ \tan{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 - \tan^2{\frac{\alpha}{2} } } \end{array} ~} \quad \text{ or } \quad \boxed{~ \begin{array}{rcl} \sin{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 + \tan^2{\alpha } } \\\\ \cos{(2\alpha)} &=& \dfrac{ 1 - \tan^2{\alpha } } { 1 + \tan^2{\alpha } }\\\\ \tan{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 - \tan^2{\alpha } } \end{array} ~} \end{array}\)
\(\begin{array}{rcl} \sin{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 + \tan^2{x } } \\ \sin{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 + ( -\frac12 )^2 } \\ \sin{(2x)} &=& \dfrac{ -1 } { 1 + \frac14 } \\ \mathbf{\sin{(2x)}} &\mathbf{=}& \mathbf{-\frac45}\\ \\ \cos{(2x)} &=& \dfrac{ 1 - \tan^2{x } } { 1 + \tan^2{x } }\\ \cos{(2x)} &=& \dfrac{ 1 - \frac14 } { 1 + \frac14 }\\ \cos{(2x)} &=& \frac34 \cdot \frac45 \\ \mathbf{\cos{(2x)}} &\mathbf{=}& \mathbf{\frac35} \\ \\ \tan{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 - \tan^2{x } } \\ \tan{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 - ( -\frac12 )^2 }\\ \tan{(2x)} &=& \dfrac{ -1 } { 1 - \frac14 }\\ \mathbf{\tan{(2x)}} &\mathbf{=}& \mathbf{- \dfrac43}\\ \end{array}\)
+118723 Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0
If a right angles triangle has the two short sides of 1 and 2 then the hypotenuse must be sqrt5
since tan is neg the angle is in the 2nd or 4th quads and since cos is pos, x must be in the 4th quad.
tan x = −1/2 cosx= +2/sqrt5 sinx= -1/sqrt5
sin2x= 2sinxcosx = 2* -1/sqrt5 * 2/sqrt5 = -4/5
cos2x= cos^2x-sin^2x = 4/5-1/5 = 3/5
I did it in my head so I hope I didn't do anything stupid. You are capable of checking yourself.
Any question, just ask. :)
+26400 Find sin 2x, cos 2x, and tan 2x from the given information. tan x = −1/2 , cos x > 0
In trigonometry, the tangent half-angle formulas relate the tangent of one half of an angle to trigonometric functions of the entire angle. They are as follows:
\(\begin{array}{rcl} \boxed{~ \begin{array}{rcl} \sin{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } } \\\\ \cos{(\alpha)} &=& \dfrac{ 1 - \tan^2{\frac{\alpha}{2} } } { 1 + \tan^2{\frac{\alpha}{2} } }\\\\ \tan{(\alpha)} &=& \dfrac{ 2\cdot \tan{\frac{\alpha}{2} } } { 1 - \tan^2{\frac{\alpha}{2} } } \end{array} ~} \quad \text{ or } \quad \boxed{~ \begin{array}{rcl} \sin{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 + \tan^2{\alpha } } \\\\ \cos{(2\alpha)} &=& \dfrac{ 1 - \tan^2{\alpha } } { 1 + \tan^2{\alpha } }\\\\ \tan{(2\alpha)} &=& \dfrac{ 2\cdot \tan{\alpha } } { 1 - \tan^2{\alpha } } \end{array} ~} \end{array}\)
\(\begin{array}{rcl} \sin{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 + \tan^2{x } } \\ \sin{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 + ( -\frac12 )^2 } \\ \sin{(2x)} &=& \dfrac{ -1 } { 1 + \frac14 } \\ \mathbf{\sin{(2x)}} &\mathbf{=}& \mathbf{-\frac45}\\ \\ \cos{(2x)} &=& \dfrac{ 1 - \tan^2{x } } { 1 + \tan^2{x } }\\ \cos{(2x)} &=& \dfrac{ 1 - \frac14 } { 1 + \frac14 }\\ \cos{(2x)} &=& \frac34 \cdot \frac45 \\ \mathbf{\cos{(2x)}} &\mathbf{=}& \mathbf{\frac35} \\ \\ \tan{(2x)} &=& \dfrac{ 2\cdot \tan{x } } { 1 - \tan^2{x } } \\ \tan{(2x)} &=& \dfrac{ 2\cdot ( -\frac12 ) } { 1 - ( -\frac12 )^2 }\\ \tan{(2x)} &=& \dfrac{ -1 } { 1 - \frac14 }\\ \mathbf{\tan{(2x)}} &\mathbf{=}& \mathbf{- \dfrac43}\\ \end{array}\)
