Question

please answer 1,2 and 3!

1. 2.

Verify that the equation is an identity. (Hint: sin 2x = sin (x + x)) sin 2x = 2 sin x cOS X Substitute 2x = x + x and apply the sine of a sum identity. sin 2x = sin (x + x) (Do not simplify.)

3.

Answer 1

On (We have to verify that the equation is an identity. Sinax = 2sins.cosx. We know that sin(A+B) = sinA.cosB + COSA.sinB ---- C LHS sin (2x) sin (x+2) 2x = x + x sinu.cosx + cosx- sink (Applying formula (1) Keeping A A=B=x 2 sink.cosx = RHS LHS = RHS proved. Hence

Date Page are Q(2) Given cos (s) : 3 and sin(t) = -12 73 also S'& t in Quadrant IV, in which sin(s) is negative cost is tve. sin (s) = sin's tops - C C Vi - Cos?6) с c = sin(s) = Cos SE 5 (Given) - sin (s) = - 9 25 * sin(s) = -4. 125-9 25 16 25 & tan (S) = -4 sincs) Cos(s) => tanº) = -5 again, cost = Vi-sin't sin't + cos²t 1 cost= 2 -12 13 sint = (Given) 13 cost = 144 + 169 169-144 169 cost 25 169 독 13 and tan(t) = sin(t) cos ( 613) 12 13 -12 -12 - tanit) -

Date Page Q(2) Solution continued :- Now, we have to find @ sin(S#t) = sin(). Cost + casts). sinit) Lusing Sin(A+B) - Sina.losbe tasA.sinB + 1-26 65 - 46 6 5 Ans 3) Sin (S+ U) = 1-9) + (3):6) (-20)+(-26) » Sin (s+t) = 32 165 & sin(s+t) (6) tan (SIL) tan(s) + tan (t) using: fan{A+B) = =tan(s+t) = 65 - tan/s+t) lis (-4).(-12) fandt fanß 1- fans). tant) latanA tanB) -20 +(-30) 15 115- 48 IS is - tan(s+t) こ 456 1-56 T5 -33 IS X to (+33) 56 33 ts .:: tom (5+t) 86 Ans 33 (c) Given s & t are in Qudrant (IV) le 3R 2 CS L 27 34 ct 27 2 (34 +36) < (s+ + < (2x +29) 34 < (S+t) 2-4* = (stt) is in 3rd & 4th Quad dant-

Page (3) Given Cos(s) = 3 and cos(t) = - 8 Sandt 17 in Quadrant , in which sin (s) is tve and sin(t) is also positive. sin?5) + cos? (s) = ..Shnts) = Vi-carlos (-sin?c) + 60() = 1) $55) = -9) = sin(s) = 6 25 4 5 and tan (s) = sin(s) Cos(s) 4 xs -4 = tan (5) = tan (5) is -ve in Ind Qudrant, it meems o the value of tan (3), which I Calculated is true) In the same way sin(t) = I-cost = sin(t) = 64 289 1289-64) 289 3 17 = sin(t) = 225 289 and tan (t) = sin(t) Cost) - tan(t) = -15 (4) (음) n -8

Date Page - 8.3) Solution is continued: Now, we have to find (a) Sin (s+t) = sin (S). Costt) + 39). Sintt) =) Sin (5+1) = Ⓡ ). ( + ) Fisin{s++) = :) + Ghis (32) (45) (-32) + (15) 85 5) Sin (s+t) =) sin(st t). (6) tan(s+t) + 85 85 -77 85 tan(s) + tan(t) - tan(s). tan(t) -15 + → tan (s+t) = 4 -IS 병 [132) +, 445) ] ) tan(s+t) 24 60 24. –77 24 (24-60 24 60 tancs+t) = (-77 24 (-36 24 77 36 +36 je ce tan (5+1) = 77 Ars Given S & t are in Quadrant (I) I t tt (5 ++)? = 6 +5}<(s+t) <(+a) it means (tt) lies => 7 < (5+t) < 27=> in god e 4h Quadrant KIN CS LT &