Use the cofunction identities to evaluate the expression below without the aid of a calculator. $\sin ^ { 2 } 62 ^ { \circ } + \sin ^ { 2 } 30 ^ { \circ } + \sin ^ { 2 } 28 ^ { \circ } + \sin ^ { 2 } 60 ^ { \circ }$
A) 1
B) 2
C) $- 1$
D) 0
E) $\frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

Using the cofunction identities, we have:
$$\sin ^{2}(62^{\circ})=\cos ^{2}(28^{\circ})$$
$$\sin ^{2}(30^{\circ})=\cos ^{2}(60^{\circ})$$
So the expression becomes:
$$\cos ^{2}(28^{\circ})+\cos ^{2}(60^{\circ})+\sin ^{2}(28^{\circ})+\cos ^{2}(60^{\circ})$$
$$=2\cos ^{2}(60^{\circ})+2\cos ^{2}(28^{\circ})$$
Since $\cos(60^{\circ})=\frac{1}{2}$, we have:
$$2\cos ^{2}(60^{\circ})+2\cos ^{2}(28^{\circ})=2\left(\frac{1}{2}ight)^2+2\cos^2(28^{\circ})=1+\cos^2(28^{\circ})$$
Using the identity $\sin^2x+\cos^2x=1$, we have:
$$1+\cos^2(28^{\circ})=1+\left(1-\sin^2(28^{\circ})ight)=2-\sin^2(28^{\circ})$$
Now, using the cofunction identity $\sin(90^{\circ}-x)=\cos(x)$, we have:
$$\sin(90^{\circ}-28^{\circ})=\sin(62^{\circ})=\cos(28^{\circ})$$
Therefore:
$$2-\sin^2(28^{\circ})=2-\cos^2(62^{\circ})=\sin^2(62^{\circ})+\cos^2(62^{\circ})-1=1-1=\boxed{	extbf{(D) }0}$$

Use the Cofunction Identities to Evaluate the Expression Below Without sin⁡262∘+sin⁡230∘+sin⁡228∘+sin⁡260∘\sin ^ { 2 } 62 ^ { \circ } + \sin ^ { 2 } 30 ^ { \circ } + \sin ^ { 2 } 28 ^ { \circ } + \sin ^ { 2 } 60 ^ { \circ }sin262∘+sin230∘+sin228∘+sin260∘

Use the Cofunction Identities to Evaluate the Expression Below Without $\sin ^ { 2 } 62 ^ { \circ } + \sin ^ { 2 } 30 ^ { \circ } + \sin ^ { 2 } 28 ^ { \circ } + \sin ^ { 2 } 60 ^ { \circ }$