Question

(iii) cos2 45°- sin 45° | HA
TT ...​

Answer 1

Answer:

Show that cos^2 (45°)-sin^2 (15°) = √3/4?

cos245∘−sin215∘

=cos245∘−sin2(45∘−30∘)(LHS)

We know that sin(A−B)=sinA⋅cosB−cosA⋅sinB

Therefore, LHS =cos245∘−(sin45∘⋅cos30∘−cos45∘⋅sin30∘)2

=12−(12–√×3–√2−12–√×12)2

=12−(3–√22–√−122–√)2

=12−(3–√−122–√)

= 12−((3–√−1)28)

=12−(3–√−1)28

=12−4−23–√8

=12−2−3–√4

=2−(2+3–√)4

=3–√4= RHS.

Hence, the problem is solved.

Answer

Sin (15°) = Sin (45°— 30°)

Now according to the formula

Sin(A— B) = sinAcosB — cosAsinB (HERE A AND B ARE TWO DIFFERENT ANGLES)

So by applying it you get

Sin (15°) = sin(45°)cos(30) — cos(45°)sin(30°)

= 1/root 2 × root3/2 — 1/root2 × 1/2

= root3/2 root2 — 1/2 root2

= root3 — 1/ 2 root2

So by squaring this

We get

4– 2 root3/ 8

Now put the values in the formula

Cos^2 (45°) — sin^2(15°)

= 1/2 — ( 1/2 — root3/4)

= root3/4