Question 127

The value of the expression $$Sin^{2}1^\circ+Sin^{2}11^\circ+Sin^{2}21^\circ+Sin^{2}31^\circ+Sin^{2}41^\circ+Sin^{2}45^\circ$$

$$+Sin^{2}49^\circ+Sin^{2}59^\circ+Sin^{2}69^\circ+Sin^{2}79^\circ+Sin^{2}89^\circ$$

Solution

Expression : $$Sin^{2}1^\circ+Sin^{2}11^\circ+Sin^{2}21^\circ+Sin^{2}31^\circ+Sin^{2}41^\circ+Sin^{2}45^\circ$$ $$+Sin^{2}49^\circ+Sin^{2}59^\circ+Sin^{2}69^\circ+Sin^{2}79^\circ+Sin^{2}89^\circ$$

= $$(Sin^{2}1^\circ+Sin^{2}89^\circ)+(Sin^{2}11^\circ+Sin^{2}79^\circ)+(Sin^{2}21^\circ+Sin^{2}69^\circ)+(Sin^{2}31^\circ+Sin^{2}59^\circ)+$$ $$(Sin^{2}41^\circ+Sin^{2}49^\circ)+(Sin^{2}45^\circ)$$

= $$[Sin^{2}1^\circ+Sin^{2}(90^\circ-1^\circ)]+[Sin^{2}11^\circ+Sin^{2}(90^\circ-11^\circ)]+[Sin^{2}21^\circ+Sin^{2}(90^\circ-21^\circ)]+$$ $$[Sin^{2}31^\circ+Sin^{2}(90^\circ-31^\circ)]+[Sin^{2}41^\circ+Sin^{2}(90^\circ-41^\circ)]+[Sin^{2}45^\circ]$$

Using, $$sin(90^\circ-\ \theta)=cos\ \theta$$

= $$(Sin^{2}1^\circ+Cos^{2}1^\circ)+(Sin^{2}11^\circ+Cos^{2}11^\circ)+(Sin^{2}21^\circ+Cos^{2}21^\circ)+(Sin^{2}31^\circ+Cos^{2}31^\circ)+$$ $$(Sin^{2}41^\circ+Cos^{2}41^\circ)+(Sin^{2}45^\circ)$$

Using, $$sin^2\ \theta+cos^2\ \theta=1$$

= $$(1+1+1+1+1)+(\frac{1}{\sqrt2})^2$$

= $$5+\frac{1}{2}=5\frac{1}{2}$$

=> Ans - (B)


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