What is the value of cot^2(135)-sin(210)+5cos^2(225) ?

The value of cot^2(135)-sin(210)+5cos^2(225) is 4

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cot^2(135)-sin(210)+5cos^2(225)

cot^{2} (13 5^{\circ}) - sin (21 0^{\circ}) + 5 cos^{2} (22 5^{\circ})

4

Solution steps

cot^{2} (13 5^{\circ}) - sin (21 0^{\circ}) + 5 cos^{2} (22 5^{\circ})

Rewrite using trig identities:

cos^{2} (22 5^{\circ}) = \frac{1 + cos ( 9 0 ^{\circ} )}{2}

= cot^{2} (13 5^{\circ}) - sin (21 0^{\circ}) + 5 \cdot \frac{1 + cos ( 9 0 ^{\circ} )}{2}

Simplify

= \frac{2 cot ^{2} ( 13 5 ^{\circ} ) - 2 sin ( 21 0 ^{\circ} ) + 5 ( 1 + cos ( 9 0 ^{\circ} ) )}{2}

Use the following trivial identity:

cot (13 5^{\circ}) = - 1

Rewrite using trig identities:

sin (21 0^{\circ}) = - \frac{1}{2}

Use the following trivial identity:

cos (9 0^{\circ}) = 0

= \frac{2 ( - 1 ) ^{2} - 2 ( - \frac{1}{2} ) + 5 ( 1 + 0 )}{2}

Simplify

\frac{2 ( - 1 ) ^{2} - 2 ( - \frac{1}{2} ) + 5 ( 1 + 0 )}{2} : 4

= 4

sin (1)

2 sin^{2} (θ) = 1 - sin (θ)

sin (2 x) - sin (x) = 0

sin (54 0^{\circ})

arccos (0.5)

What is the value of cot^2(135)-sin(210)+5cos^2(225) ?
The value of cot^2(135)-sin(210)+5cos^2(225) is 4