In any triangle ABC prove that cos²A+cos²(60+A)+cos²(60-A)=3/2
Answers
Answer:
{cos}^{2} A + { \cos}^{2} (A + 60) + {cos}^{2} ( A - 60) = \frac{3}{2} cos
2
A+cos
2
(A+60)+cos
2
(A−60)=
2
3
Using the formula
\small{\cos( \alpha + \beta ) = \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta )} cos(α+β)=cos(α)cos(β)−sin(α)sin(β)
and
\small{\cos( \alpha - \beta ) = \cos( \alpha ) \cos( \beta ) + \sin( \alpha ) \sin( \beta ) }cos(α−β)=cos(α)cos(β)+sin(α)sin(β)
\small{cos}^{2} A + {cos}^{2} A \: {cos}^{2} 60 \degree - {sin}^{2} A \: {sin}^{2} 60 \degree + {cos}^{2}A \: {cos}^{2} 60 \degree + {sin}^{2} A \: {sin}^{2} 60 \degree \: = \frac{3}{2} cos
2
A+cos
2
Acos
2
60°−sin
2
Asin
2
60°+cos
2
Acos
2
60°+sin
2
Asin
2
60°=
2
3
\small {cos}^{2} A + \frac{1}{4} {cos}^{2} A + \frac{1}{4} {cos}^{2} A = \frac{3}{2} cos
2
A+
4
1
cos
2
A+
4
1
cos
2
A=
2
3
{cos}^{2} A(1 + \frac{1}{4} + \frac{1}{4} ) = \frac{3}{2} cos
2
A(1+
4
1
+
4
1
)=
2
3
\frac{3}{2} {cos}^{2} A = \frac{3}{2}
2
3
cos
2
A=
2
3
{cos}^{2} A = 1 \implies \: cos \: A = + - 1cos
2
A=1⟹cosA=+−1
A = \color{green}90 \degree \: or \: 270 \degreeA=90°or270°
Hope it helps ✌️ ✌️
Step-by-step explanation: