Trignometric Ratios and Identities |
| Math: Trignometry Chapter
Some nice manipulations: 1) Why ? = cos2
sin2 2sin
cos = 1 2sin cos![]() = 1 sin2![]() 2) Why ? = ![]() 3) cos Asin A = ![]() Why ! cos A sin A = ![]() = ![]() = ![]() Expressing sin in terms of sin
A: =
= ![]() The ambiguity of sign is removed from following figure. ![]() Dumb Question: Where does this diagram come from ? Ans:- So, if is in I st or IInd
quadrent then is + ve. So, A/2 lie from 2n
for to be + veSimilarily for ![]() Illustration 7: If cos 250 + sin 25 0 = P , then find value of cos 500 in trems of P ? Ans:- cos500 = cos2 25 0 - sin2 250 = (cos 250 + sin250)(cos 250 - sin250) P(cos250 - sin250) Also, (cos250 - sin250)2 + (cos250 - sin250)2 = 1 + 1 cos250 - sin250 = +
(+ve sign as cos 250 > sin 250) cos500 = ![]() The qreatest and least valume of expression (a sin
+ b cos )-
a sin ![]() Why? Let a = r cos b = r sin so that r = ![]() So, a sin
+ cos )
= r sin ( + )Now sin ( + has
minimum and maximum value as + 1 and - 1 esppectively.So, - r rsin ( + )
r So, -
a sin ![]() Illustration 8: Find the minimum and maximum value of 6 sin x cos x + 4cos2 2x ? Ans:- 6 sin x cos x + 4cos2 2x = (2sin x cos x ) + 4cos 2x = 3 sin 2x + 4cos 2x . - 3sin 2x + 4cos2x
=> Minimum value of 6sin x cos x + 4 cos 2x is - 5 and maximum value is 5. Sum of sine and cosine senies when angles are in AP . (1) sin + sin
= Why ? ![]() 2sin 2sin ![]() 2sin ![]() By adding these n lines we have. 2sin = 2sin S = ![]() (2) = ![]() Illustration 9: Find sum of sin -............+
0 n terms. Ans:- Now, sin sin sin ![]() ........................................................ Hence the series is sin ............ = = ![]() |


= cos2
sin2
2sin





in terms of sin
A:

is in I st or IInd
quadrent
is + ve.
for
to be + ve

a sin
3sin 2x + 4cos2x
=





-............+
0 n terms. Ans:- Now, sin

............



