Mathematics: Find the value of a. sin 120 degrees b. tan 150 degrees c. cos (-135 degrees) d. sec 300 degrees e. sin 240 - cos 330 degrees

Find the value of a. sin 120 degrees b. tan 150

Ответ:

aud50

04.06.2020

$A.\ sin120^o=sin(2\cdot60^o)=2sin60^ocos60^o=2\cdot\frac{\sqrt3}{2}\cdot\frac{1}{2}=\frac{\sqrt3}{2}\\\\B.\ tan150^o=tan(90^o+60^o)=-ctg60^o=-\frac{\sqrt3}{3}\\\\C.\ cos(-135^o)=cos135^o=cos(90^o+45^o)=-sin45^o=-\frac{\sqrt2}{2}\\\\D.\ sec300^o=\frac{1}{cos300^o}=\frac{1}{cos(360^o-60^o)}=\frac{1}{cos60^o}=\frac{1}{\frac{1}{2}}=2\\\\E.\ sin240^o-cos330^o=sin(180^o+60^o)-cos(360^o-30^o)\\\\=-sin60^o-cos30^o=-\frac{\sqrt3}{2}-\frac{\sqrt3}{2}=-\frac{2\sqrt3}{2}=-\sqrt3$

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Ответ:

huddyxo

09.10.2020

Mathematics

A) sin 120=sin (180-60)=sin 60= $\frac{ \sqrt{3} }{2}$ (It lies on second quadrant where sin is +ve)
B) tan (150)=tan(180-30)=-tan30= $\frac{-1}{ \sqrt{3} }$ (It lies on second quadrant where tan is -ve)
C) cos (-135)=cos (135)=cos (180-45)=-cos (45)= $\frac{-1}{ \sqrt{2} }$ (It lies on second quadrant where cos is -ve)
D) sec (300)=sec (360-60)=sec(60)=2 (It lies on fourth quadrant where sec is +ve)
E) sin (240)=sin(180-(-60))=sin(-60)=-sin (60)= $\frac{- \sqrt{3} }{2}$
(It lies on third quadrant where sin is -ve)
F) cos (330)=cos(360-30)=cos(30)= $\frac{ \sqrt{3} }{2}$
(It lies on fourth quadrant where cos is +ve)
sin (240)- cos (330)= $- \sqrt{3}$

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Ответ:

pal23

14.03.2023

Mathematics

First I would need to know if x and y represent a number

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Find the value of a. sin 120 degrees b. tan 150 degrees c. cos (-135 degrees) d. sec 300 degrees e. sin 240 - cos 330 degrees

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