Question

Value of sin(nπ-x) ??

Answer 1

Answer:

ans= sin(n¶-x) =sinx

as trigono. fun(¶-x) =That trigono. fun.

+ve value only for sin fun.

Answer 2

Answer:

Sin (nπ + x) = (-1)^n sin x

Proof by induction method

For n= 1

Sin(π + x) = sin π cos x + cos π sin x

=>Sin(π+x) = 0-sin x= (-1)^1 sin x ……..(1)

for n= k ,let

sin(kπ +x) = (-1)^k sin x ……….(2)

Now for n= k+1

Sin((k+1)π + x) = sin ( kπ +(π+x) )

=>Sin((k+1)π + x ) = (-1)^k sin (π+x)

[ we have assumed that sin (kπ +x) = (-1)^k sinx]

Now we know that from equation 1

sin (π+x) = -sin x

Therefore Sin ((k+1)π + x) = (-1)^k *(-sin x) = (-1)^(k+1) sin x

Hence in general we can say Sin (nπ + x ) = (-1)^n sin x

So if n is odd sin (nπ + x) = -sin x

And if n is even sin (nπ + x) = sin x

Hope it helps !!!

Step-by-step explanation: