2016年8月15日 · When #theta# becomes equal to #pi/2# then adjacent will vanish and opposite will coincide with the hypotenuse. This means when #theta=pi/2" then # "opposite"="hypotenuse"# So then

2015年4月16日 · cos x With pi/2 add to any angle measure, sin changes to cos and vice- versa. Hence It would change to cosine and since the angle measure falls in the second quadrant, hence sin(x+pi/2) would be positive. Alternatively sin(x+pi/2)= sin x cos pi/2 + cos x sinpi/2. Since cos pi/2 is 0 and sinpi/2 is 1, it would be equal to cosx

2016年3月15日 · How do you apply the fundamental identities to values of #theta# and show that they are true

2016年8月16日 · If sinA=4/5 and cosB= -5/13, where A belongs to QI and B belongs to QIII, then find sin(A+B).... See all questions in Sum and Difference Identities Impact of this question

2016年4月9日 · The right hand side should be - cos x Expand sin(x-pi/2) using sin(a-b)=sin a cos b-cos a sin b. Prove sin (x-pi/2 )=-cos x.

2015年4月23日 · for the "true" proof you need to use matrice, but this is acceptable : sin(a+b) = sin(a)cos(b)+cos(a)sin(b) sin(pi/2+x) = sin(pi/2)*cos(x)+cos(pi/2)*sin(x) sin(pi/2) = 1 cos(pi/2) = 0 So we have : sin(pi/2+x) = cos(x) Since this answer is very usefull for student here the full demonstration to obtain sin(a+b) = sin(a)cos(b)+cos(a)sin(b) (do not read this if you are not …

2017年2月2日 · sin(a-pi/2)=-m We can use two formulas here. One sin(-A)=-sinA and sin(pi/2-A)=cosA Hence sin(a-pi/2)=sin(-(pi/2-a)) = -sin(pi/2-a)=-cosa=-m

2015年5月15日 · Then the side opposite the angle #x# will have length #sin x# and the side opposite the angle #(pi/2 - x)# will have length #sin (pi/2 - x)#. By Pythagoras theorem, the sum of the squares of the lengths of these sides is equal to the square of the length of the hypotenuse. So #sin^2 x + sin^2 (pi/2 - x) = 1^2 = 1#

2018年3月2日 · There are 2 ways: 1. Use trig identity: sin (a + b) = sin a.cos b + sin b.cos a In this case sin (x + pi/2) = sin x.cos (pi/2) + sin (pi/2).cos x. Because cos (pi/2) = 0; and sin (x/2) = 1, therefor, sin (x + pi/2) = cos x 2. By the unit circle.

2016年1月20日 · -pi/2 If you want to know how please follow. sin(-pi/2) = -sin(pi/2) since sin(x) is a odd function. sin(-pi/2) = -1 since sin(pi/2) = 1 arcsin(sin(-pi/2)) = arcsin(-1) Now comes the range of arcsin(x) The range of arcsin(x) is [-pi/2,pi/2] So arcsin(-1) would give -pi/2 Therefore, arcsin(sin(-pi/2)) = -pi/2