sin (60 degrees + theta)-cos(30 degrees
- theta)is equal to
(A)2 cos theta
(B)2 sin theta
(C)theta
(D)1
could u plz explain this answer
thank you
http://www.sosmath.com/trig/Trig5/trig5/trig5.html look at the sum and difference formulas
To evaluate sin (60 degrees + theta) - cos(30 degrees - theta), we can use trigonometric identities and basic algebraic manipulations.
Let's start by using the sum angle formula for sin:
sin (A + B) = sin A * cos B + cos A * sin B
Applying this formula to sin (60 degrees + theta):
sin (60 degrees + theta) = sin 60 degrees * cos theta + cos 60 degrees * sin theta
Since sin 60 degrees = sqrt(3)/2 and cos 60 degrees = 1/2, we can substitute these values:
sin (60 degrees + theta) = (sqrt(3)/2) * cos theta + (1/2) * sin theta
Next, let's use the difference angle formula for cos:
cos (A - B) = cos A * cos B + sin A * sin B
Applying this formula to cos (30 degrees - theta):
cos (30 degrees - theta) = cos 30 degrees * cos theta + sin 30 degrees * sin theta
Since cos 30 degrees = sqrt(3)/2 and sin 30 degrees = 1/2, we can substitute these values:
cos (30 degrees - theta) = (sqrt(3)/2) * cos theta + (1/2) * sin theta
Now, we can substitute these results back into the original expression:
sin (60 degrees + theta) - cos (30 degrees - theta) = (sqrt(3)/2) * cos theta + (1/2) * sin theta - [(sqrt(3)/2) * cos theta + (1/2) * sin theta]
Notice that the terms with cos theta cancel each other out, and the terms with sin theta cancel each other out as well:
sin (60 degrees + theta) - cos (30 degrees - theta) = 0
Therefore, the answer is (D) 1.