Question

4 sin square 60 degree + 3 x square 30 degree minus 8 science square 45 degree minus cos square 45 degree

Answer 1

Hello mate!!

Given

4sin²60° + 3tan²30° - 8sin45° . cos45°

4 . ( √3 / 2 )² + 3 ( 1 /√3 )² - 8 ( 1 /√2 ) . ( 1 /√2 )

= 4 × 3 / 4 + 3 × 1 / 3 - 8 × 1 / 2

= 3 + 1 - 4

= 3 - 3



Hope it helps you!!

Answer 2

$4\sin^2 60^\circ+ 3\tan^2 30^\circ - 8\sin^2 45^\circ-\cos^2 45^\circ=-\frac{1}{2}$

Step-by-step explanation:

Given : Expression $4\sin^2 60^\circ+ 3\tan^2 30^\circ - 8\sin^2 45^\circ-\cos^2 45^\circ$

To find : Solve the expression ?

Solution :

$4\sin^2 60^\circ+ 3\tan^2 30^\circ - 8\sin^2 45^\circ-\cos^2 45^\circ$

Using trigonometric values,

$\sin 60^\circ=\frac{\sqrt3}{2}$

$\sin 45^\circ=\frac{1}{\sqrt2}$

$\cos 45^\circ=\frac{1}{\sqrt2}$

$\tan 30^\circ=\frac{1}{\sqrt3}$

Substitute the values,

$=4(\frac{\sqrt3}{2})^2+ 3(\frac{1}{\sqrt3})^2 - 8(\frac{1}{\sqrt2})^2-(\frac{1}{\sqrt2})^2$

$=4(\frac{3}{4})+ 3(\frac{1}{3})- 8(\frac{1}{2})-(\frac{1}{2})$

$=3+1-4-(\frac{1}{2})$

$=-\frac{1}{2}$

Therefore, $4\sin^2 60^\circ+ 3\tan^2 30^\circ - 8\sin^2 45^\circ-\cos^2 45^\circ=-\frac{1}{2}$

#Learn more

4 cot square 60 degree + sin square 30 degree minus 2 sin squared 45 degree by sin square 60 degree + cos square 45 degree

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