Question

The value of ${\text{sin}}^2{\text{5°+sin}}^2{\text{10°+sin}}^2{\text{15°+⋯⋯⋯⋯⋯⋯⋯+sin}}^2{\text{85°+sin}}^2\text{90°}$ is

• A. $7$
• B. $8$
• C. $9.5$
• D. $10$

Answer 1

The correct option is C

$9.5$

We know that ${\text{sin(90-x)=cosx,sin}}^2{\text{(90-x)=cos}}^2\text{x}$

And also we have ${\text{sin}}^2{\text{x+cos}}^2\text{x=1}$

$$\begin{gathered} \Rightarrow si{n}^{2}5°+si{n}^{2}10°+si{n}^{2}15°+\cdots \cdots \cdots \cdots \cdots +si{n}^{2}85°+si{n}^{2}90° \\ =si{n}^{2}5°+si{n}^{2}10°+si{n}^{2}15°+\cdots \cdots \cdots \cdots \cdots +si{n}^2(90-5)°+si{n}^{2}90° \\ ={\sin }^{2}5°+{\sin }^{2}10°+{\sin }^{2}15°+\cdots \cdots \cdots \cdots \cdots +{\cos }^{2}10°+{\cos }^{2}5°+{\sin }^{2}90° \\ =({\sin }^{2}5°+{\cos }^{2}5°)+({\sin }^{2}10°+{\cos }^{2}10°)+({\sin }^{2}15°+{\cos }^{2}15°)+({\sin }^{2}20°+{\cos }^{2}20°)+({\sin }^{2}25°+{\cos }^{2}25°)+({\sin }^{2}30°+{\cos }^{2}30°)+({\sin }^{2}35°+{\cos }^{2}35°)+({\sin }^{2}40°+{\cos }^{2}40°)+{\sin }^{2}45°+{\sin }^{2}90° \\ =1+1+1+1+1+1+1+1+\frac{1}{2}+1(weknowthat\sin 45°=\frac{1}{\sqrt{2}},\sin 90°=1) \\ =9+\frac{1}{2}=\frac{18+1}{2}=\frac{19}{2} \\ =9.5 \end{gathered}$$

Hence option(C) is the correct answer and all other options are incorrect answer .