Oct 1, 2015 · At #270^@# for a unit radius circle #color(white)("XXX")sin(270^@) = ("opposite")/("hypotenuse")# #color(white)("XXXXXXXXX")=y/sqrt(x^2+y^2)#

Jul 14, 2018 · Answer: The answer is 0. Step-by-step explanation: Given, ABCD is a cyclic quadrilateral. We know that, The sum of the opposite angles of a cyclic quadrilateral is 180°,

Mar 13, 2019 · cos 180 = cos(270-90) cos 180 = -sin 90 cos(270-a) = -sin a, cos 180 = -1. Advertisement riya666619 riya666619

Apr 30, 2015 · cos 240 = (-1)/2 240 is in the IIIrd quadrant hence it would be negative. This can be determined as cos 240 = cos (180+60) = -cos 60 = (-1)/2

Feb 15, 2016 · From the given #2=r*cos (theta+180)# #2=r*[cos theta cos 180-sin theta sin 180]# #2=r*[cos theta(-1)-sin theta (0)]#

Jan 18, 2025 · To find the value of sin 150° - cos 180°: *Step 1: Find the value of sin 150°* Using the trigonometric identity sin (180° - x) = sin x, we can rewrite sin 150° as sin (180° - 30°) = sin …

Mar 20, 2018 · You can use this altered Pythagorean identity: #sin^2=1-cos^2x# Now here's the actual problem. The strategy is to get everything in terms of #cosx#, then factor it like a quadratic.

Feb 11, 2018 · a+b+c=180^@, LHS=cos(b+c-a)+cos(c+a-b)+cos(a+b-c) =cos(180-2a)+cos(180-2b)+cos(180-2c) =-cos(2a)-cos(2b)-cos(2c) =-(2cos^2(a)-1)-(cos(2b)+cos(2c)) =1-2cos^2(a) …

Dec 26, 2019 · Then, cos A + cos B = cos A + cos (180° - A)= cos A - cos A [because cos (180° - x) = - cos x]= 0 harishramayanam harishramayanam 26.12.2019

Jun 29, 2016 · cos(-120^o)=-1/2 cos(-120^o)=cos120^o = cos(180^o-60^o) = -cos60^o = -1/2