Question
Evaluate the following:
$\sin^230^\circ+\sin^245^\circ+\sin^260^\circ+\sin^260^\circ+\sin^290^\circ\dots(1)$
$\sin^230^\circ+\sin^245^\circ+\sin^260^\circ+\sin^260^\circ+\sin^290^\circ\dots(1)$
✓
Answer
$\sin^230^\circ+\sin^245^\circ+\sin^260^\circ+\sin^260^\circ+\sin^290^\circ\dots(1)$
By trigonometric ratios we have
$\sin30^\circ=\frac{1}{2}\ \ \ \ \sin45^\circ=\frac{1}{\sqrt2}$
$\sin60^\circ=\frac{\sqrt3}{2}\ \ \ \ \sin90^\circ=1$
By substituting above values in (i), we get
$=\Big[\frac{1}{2}\Big]^2+\Big[\frac{1}{\sqrt2}\Big]^2+\Big[\frac{\sqrt3}{2}\Big]^2+[1]^2$
$=\frac{1}{4}+\frac{1}{2}+\frac{3}{2}+1\Rightarrow\frac{1+3}{4}+\frac{1+2}{2}$
$\Rightarrow1+\frac{3}{2}=\frac{2+3}{2}=\frac{5}{2}$
By trigonometric ratios we have
$\sin30^\circ=\frac{1}{2}\ \ \ \ \sin45^\circ=\frac{1}{\sqrt2}$
$\sin60^\circ=\frac{\sqrt3}{2}\ \ \ \ \sin90^\circ=1$
By substituting above values in (i), we get
$=\Big[\frac{1}{2}\Big]^2+\Big[\frac{1}{\sqrt2}\Big]^2+\Big[\frac{\sqrt3}{2}\Big]^2+[1]^2$
$=\frac{1}{4}+\frac{1}{2}+\frac{3}{2}+1\Rightarrow\frac{1+3}{4}+\frac{1+2}{2}$
$\Rightarrow1+\frac{3}{2}=\frac{2+3}{2}=\frac{5}{2}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Seg AB and seg AD are the chords of the circle. C is a point on tangent of the circle at point A. If m(arc APB) 80° = and $\angle$BAD = 30°. Then find (i) $\angle$BAC (ii) m(arc BQD).
→The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Write the coordinates of a point on x-axis which is equidistant from the points $(-3, 4)$ and $(2, 5)$.
→In the adjoining figure, $\angle PQR =90^{\circ}$ $\operatorname{seg} QS \perp$ side $PR , PS =4, PQ =6$. Find $x, y$ and $z$.
→Decide whether 301 is term of given sequence 5,11,17,23,.......
Activity :- Here, d = ____ therefore this sequence is an A.P.
a= 5, d= ____
Let nth term of this A.P. be 301.
$t _{ n }$ = a + (n-1) ____
301 = 5 + (n-1)×6
301 = 6n-1
n =$\frac{302}{6}$ = ____
But n is not positive integer
Therefore, 301 is ____ the term of sequence 5,11,17,23.........
→Activity :- Here, d = ____ therefore this sequence is an A.P.
a= 5, d= ____
Let nth term of this A.P. be 301.
$t _{ n }$ = a + (n-1) ____
301 = 5 + (n-1)×6
301 = 6n-1
n =$\frac{302}{6}$ = ____
But n is not positive integer
Therefore, 301 is ____ the term of sequence 5,11,17,23.........
The length, breadth and height of a room are 8m 25cm, 6m 75cm and 4m 50cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Solve the following quadratic equations by factorization:
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
→$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
11, 8, 5, 2, . . . In this A.P. which term is number – 151?
Solve using formula.
x2 + 6x + 5 = 0
x2 + 6x + 5 = 0
From the top of a tower 100m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45° respectively. Find the distance between the cars. $\big[\text{Take}\sqrt{3}=1.732\big]$
→