Question

${\textbf{Subject : Mathematics}}$

${\textbf{Chapter : Trigonometric Functions }}$


Class 11​

Answer 1

cos510 × cos330

cos(90×5 + 60) + cos (360-30)

-sin60 + cos30

-√3/2 + √3/2

0

Answer 2

◦•●◉✿[ welcome to the concept of Trigonometric functions ]✿◉●•◦

we know that value of cosx is repeat after 2π or 360° .

concept:

• obtain the point X which the value of trigonometric functions to be determined.
• check whether access positive and negative, if X is negative then write f(X) = -f(y).otherwise f(X) = f(y)
• express the positive value of X in step 2 in the form of X =( nπ/2)+_b ,where b belongs to (0,π/2)
• determine the quadrant in which the determinant in Ray of the angle X lie and determine the sign of the trigonometric function in that quadrant.
• if n is step 3 is an odd function then ,sinx = +_ cos B , cos X= +- cos b, ..... similarly for other trigonometry values like tan etc.where the sign on RHS of these values will be the sign obtained in step 4.
• if n is step 3 is an even function .then sinx = +- sinB , cos X= +- cos b ..... similarity for others like tan .ved sign on RHS on these values will be the sign obtain in step 4

in yr question:

cos (510°) = cos (90×5+60)

= -sin ( 60° )

= -(√3/2) ...... (1)

cos (330° ) = cos (4× 90-30)

= cos 30

= (√3/2) ........(2)

then ,

cos (510°) + cos (330°)

=( -√3/2)+ (√3/2)

[ put values from (1) and( 2)]

= 0

.•♫•♬•[ I hope it help you ❤️ ].•♫•♬•