Question

sin30 theeta.cos3theeta.sin theeta+cos theeta  = 1−sin theeta . cos theeta


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Answer 1

Answer:

use the identity (a^3+b^3)= (a+b)(a^2+b^2-ab)

sin^3A + cos^3A÷(sinA +cos ). + sinA.cosA

(sinA+cosA)(sin^2+cos^2 -sinA.cosA)÷ (sinA + cosA)+ sinA.cosA

or( sinA^2+cosA^2 -sinA.cosA) +sinA.cosA

1-sinA.cosA+sinA.cosA

1

Step-by-step explanation:

Answer 2

Step-by-step explanation:

your answer

REFER THE ATTACHMENT.....

Let theta be equal to x.

sin ( 60°+x)

= sin 60.cos x + cos 60.sinx

=[(√3 cos x) /2 + (1 sin x) /2]

cos (30 - x)

= cos 30.cos x + sin 30.sin x

=[ (√3 cos x) /2 + (1 sin x) /2]

Therefore, sin ( 60 + x) - cos ( 30 - x)

= [(√3 cos x) /2 + (1 sin x) /2] - [(√3 cos x) /2 + (1 sin x) /2]

= 0

sin(780) = sin(780 - 720) = sin(60) = (sqrt3)/2

sind(480) = sin(480 - 360) = sin(120) = sin(60) = (sqrt3)/2

cos(120) = - cos(60) = -1/2

sin(30) = 1/2 Therefore:

sin(780) sin(480) + cos(120) sin(30) =

[(sqrt3)/2][(sqrt3)/2] + [-1/2][1/2] = (3/4) - (1/4) = 2/4 = 1/2

get FINAL ANSWER....[(√3 cos x) /2 + (1 sin x) /2] - [(√3 cos x) /2 + (1 sin x) /2 = 0