1 Answer

Xerkus · Answer 1 · 2020-01-17T18:44:24+0000

Answer:

$cos(-2835^(\circ)) = (1)/(√(2))$

$sin(-2835^(\circ)) = (1)/(√(2))$

Solution:

As per the question:

We need to find the values of:

$cos(-2835^(\circ))$

$sin(-2835^(\circ))$

Now, we know that:

$cos(- \theta) = cos\theta$

$sin(- \theta) = - sin\theta$

Also

$cos(2n\pi - \theta) = cos\theta$

$sin(2n\pi - \theta) = - sin\theta$

Now,

From the above eqn (1) and (2):

$cos(-2835^(\circ)) = cos(2835^(\circ))$

$sin(-2835^(\circ)) = - sin(2835^(\circ))$

Now the above respective values can be further calculated from eqns (3) and (4):

$cos(2(8)\pi - 45^(\circ)) = cos(45^(\circ)) = (1)/(√(2))$

$sin(2(8)\pi - 45^(\circ)) = -(- sin(45^(\circ))) = (1)/(√(2))$

where

n = 8

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