![\rm sin\dfrac{1}{2}x=\sqrt{ \dfrac{32}{54}}\\\\cos\dfrac{1}{2}x=-\sqrt{ \dfrac{2}{54} }\\\\tan\dfrac{1}{2}x=-4](https://tex.z-dn.net/?f=%5Crm%20sin%5Cdfrac%7B1%7D%7B2%7Dx%3D%5Csqrt%7B%20%5Cdfrac%7B32%7D%7B54%7D%7D%5C%5C%5C%5Ccos%5Cdfrac%7B1%7D%7B2%7Dx%3D-%5Csqrt%7B%20%5Cdfrac%7B2%7D%7B54%7D%20%7D%5C%5C%5C%5Ctan%5Cdfrac%7B1%7D%7B2%7Dx%3D-4)
<h3>Further explanation
</h3>
Trigonometry is the science of mathematics that studies the relationship between sides and angles in triangles
In a cartesian plane, the angle α can vary from 0 ° to 360 °, counterclockwise
This coordinate field is divided into 4 quadrants:
- quadrant 1: 0 ° - 90 °
- quadrant 2: 90 ° - 180 °
- quadrant 3: 180 ° - 270 °
- quadrant 4: 270 ° - 360 °
For half angle we can find it from double angle
Trigonometry Formulas for Double Angles:
sin 2A = 2 sin A cos A
cos 2A = 1 - 2 sin²A = 2cos²A-1
tan 2A = 2 tan A / 1-tan²A
then:
cos 2A = 2cos²A-1
![\rm cos\:A=2cos^2\dfrac{1}{2}A-1\\\\2cos^2\frac{1}{2}A=1+cos\:A\\\\cos^2\frac{1}{2}A=\dfrac{1+cos\:A}{2}\\\\cos\:\frac{1}{2}A=\sqrt{\dfrac{1+cos\:A}{2} }](https://tex.z-dn.net/?f=%5Crm%20cos%5C%3AA%3D2cos%5E2%5Cdfrac%7B1%7D%7B2%7DA-1%5C%5C%5C%5C2cos%5E2%5Cfrac%7B1%7D%7B2%7DA%3D1%2Bcos%5C%3AA%5C%5C%5C%5Ccos%5E2%5Cfrac%7B1%7D%7B2%7DA%3D%5Cdfrac%7B1%2Bcos%5C%3AA%7D%7B2%7D%5C%5C%5C%5Ccos%5C%3A%5Cfrac%7B1%7D%7B2%7DA%3D%5Csqrt%7B%5Cdfrac%7B1%2Bcos%5C%3AA%7D%7B2%7D%20%7D)
cos 2A = 1-2 sin²A
in the same way
![\rm sin\dfrac{1}{2}A=\sqrt{\dfrac{1-cos\:A}{2} }](https://tex.z-dn.net/?f=%5Crm%20sin%5Cdfrac%7B1%7D%7B2%7DA%3D%5Csqrt%7B%5Cdfrac%7B1-cos%5C%3AA%7D%7B2%7D%20%7D)
![\rm tan\dfrac {1} {2}A = \sqrt {\dfrac {1-cos\:A} {1+cos\:A}}](https://tex.z-dn.net/?f=%5Crm%20tan%5Cdfrac%20%7B1%7D%20%7B2%7DA%20%3D%20%5Csqrt%20%7B%5Cdfrac%20%7B1-cos%5C%3AA%7D%20%7B1%2Bcos%5C%3AA%7D%7D)
input the value : cos x = −15/17
Because x in the quadrant III (180° < x < 270°) , then
![\rm \dfrac{1}{2}x\Rightarrow quadrant\:II](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7B1%7D%7B2%7Dx%5CRightarrow%20quadrant%5C%3AII)
For quadrant I, only sin have positif value, for cos and tan have negatif value
![\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-cos\:x}{2} }](https://tex.z-dn.net/?f=%5Crm%20sin%5Cdfrac%7B1%7D%7B2%7Dx%3D%5Csqrt%7B%5Cdfrac%7B1-cos%5C%3Ax%7D%7B2%7D%20%7D)
![\rm sin\dfrac{1}{2}x=\sqrt{\dfrac{1-(-\dfrac{15}{17}) }{2} }\\\\sin\dfrac{1}{2}x=\sqrt{\dfrac{1\dfrac{15}{17} }{2} }=\sqrt{\dfrac{32}{54} }](https://tex.z-dn.net/?f=%5Crm%20sin%5Cdfrac%7B1%7D%7B2%7Dx%3D%5Csqrt%7B%5Cdfrac%7B1-%28-%5Cdfrac%7B15%7D%7B17%7D%29%20%7D%7B2%7D%20%7D%5C%5C%5C%5Csin%5Cdfrac%7B1%7D%7B2%7Dx%3D%5Csqrt%7B%5Cdfrac%7B1%5Cdfrac%7B15%7D%7B17%7D%20%7D%7B2%7D%20%7D%3D%5Csqrt%7B%5Cdfrac%7B32%7D%7B54%7D%20%7D)
![\rm cos\:\frac{1}{2}x=-\sqrt{\dfrac{1+cos\:x}{2} }\\\\cos\:\frac{1}{2}x=-\sqrt{\dfrac{1-\dfrac{15}{17} }{2} }=-\sqrt{\dfrac{2}{54} }](https://tex.z-dn.net/?f=%5Crm%20cos%5C%3A%5Cfrac%7B1%7D%7B2%7Dx%3D-%5Csqrt%7B%5Cdfrac%7B1%2Bcos%5C%3Ax%7D%7B2%7D%20%7D%5C%5C%5C%5Ccos%5C%3A%5Cfrac%7B1%7D%7B2%7Dx%3D-%5Csqrt%7B%5Cdfrac%7B1-%5Cdfrac%7B15%7D%7B17%7D%20%7D%7B2%7D%20%7D%3D-%5Csqrt%7B%5Cdfrac%7B2%7D%7B54%7D%20%7D)
![\rm tan\dfrac {1} {2}x = -\sqrt {\dfrac {1-cos\:x} {1+cos\:x}}\\\\tan\dfrac {1} {2}x = -\sqrt {\dfrac {1+\frac{15}{17} } {1-\frac{15}{17} }}\\\\tan\dfrac{1}{2}x=-\sqrt{\dfrac{\dfrac{32}{17}}{\dfrac{2}{17} } }=-\sqrt{16}=-4](https://tex.z-dn.net/?f=%5Crm%20tan%5Cdfrac%20%7B1%7D%20%7B2%7Dx%20%3D%20-%5Csqrt%20%7B%5Cdfrac%20%7B1-cos%5C%3Ax%7D%20%7B1%2Bcos%5C%3Ax%7D%7D%5C%5C%5C%5Ctan%5Cdfrac%20%7B1%7D%20%7B2%7Dx%20%3D%20-%5Csqrt%20%7B%5Cdfrac%20%7B1%2B%5Cfrac%7B15%7D%7B17%7D%20%7D%20%7B1-%5Cfrac%7B15%7D%7B17%7D%20%7D%7D%5C%5C%5C%5Ctan%5Cdfrac%7B1%7D%7B2%7Dx%3D-%5Csqrt%7B%5Cdfrac%7B%5Cdfrac%7B32%7D%7B17%7D%7D%7B%5Cdfrac%7B2%7D%7B17%7D%20%7D%20%7D%3D-%5Csqrt%7B16%7D%3D-4)
<h3>Learn more
</h3>
trigonometric ratio
brainly.com/question/9880052
Prove the converse of the Pythagorean Theorem
brainly.com/question/1597216
Use Pythagoras' theorem
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the triangle angle
brainly.com/question/1611320