2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I co related homework questions

• 2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I co

2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct?

• Proving identity (sinx+tanx)/(cosx+1)=tanx RS: (sinx+(sinx/cosx))/(cosx+1) ((sinxcosx/cosx)+(sinx/cosx))x 1/(cosx+1) sinx(cosx+1)/cosx x 1/(cosx+1) sinx/cosx = tanx RS = LS How did sinxcosx/cosx turn to sinx(cosx+1)

Proving identity(sinx+tanx)/(cosx+1)=tanxRS: (sinx+(sinx/cosx))/(cosx+1)((sinxcosx/cosx)+(sinx/cosx))x 1/(cosx+1)sinx(cosx+1)/cosx x 1/(cosx+1)sinx/cosx = tanxRS = LSHow did sinxcosx/cosx turn to sinx(cosx+1)?

• Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: RS: = 1 - 2sin^2

Prove:sin2x / 1 - cos2x = cotxMy Attempt:LS:= 2sinxcosx / - 1 - (1 - 2sin^2x)= 2sinxcosx / - 1 + 2sin^2x= cosx / sinx - 1= cosx / sinx - 1/1= cosx / sinx - sinx / sinx--Prove:2sin(x+y)sin(x-y) = cos2y - cos2xMy Attempt:RS:= 1 - 2sin^2y - 1 - 2sin^2x= 1 - 1 - 2sin^2y - 2sin^2x= -2...

• tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1-sin^2x) 2sin^2x + sinx-1=0 (2sinx+1)(sinx-1)=0 x=30 x=270 but if i plug 270 back into the original equation i get undefined beca

tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx(sinx+1)/cosx =2cosxmultiplying both sides by cosxsinx + 1 =2cos^2xsinx+1 = 2(1-sin^2x)2sin^2x + sinx-1=0(2sinx+1)(sinx-1)=0x=30 x=270but if i plug 270 back into the original equation i get undefined because tan 270 is undefined where did...

• It’s review question, I need this as soon as possible. Thank you 3) For thè diferential equation: (a) The point zo =-1 is an ordinary point. Compute the recursion formula for the coefficients of...

It’s review question, I need this as soon as possible. Thank you 3) For thè diferential equation: (a) The point zo =-1 is an ordinary point. Compute the recursion formula for the coefficients of the power series solution centered at zo- -1 and use it to compute the first th...

• Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx

Simplify #1:cscx(sin^2x+cos^2xtanx)/sinx+cosx= cscx((1)tanx)/sinx+cosx= cscxtanx/sinx+cosxIs the correct answer cscxtanx/sinx+cosx?Simplify #2:sin2x/1+cos2X= ???I'm stuck on this one. I don't know what I should do.Simplify #3:cosx-sin(90-x)sinx/cosx-cos(180-x)tanx= cosx-(sin90cos...

• Solve (sinx-1)(cosx -1/2) = 0 where 0≤x<2pi a) pi/3, pi/2, 5pi/3 b) pi/6, 5pi/6, pi c) pi/3, pi, 5pi/3 d) pi/6, pi/2, 5pi/6

Solve (sinx-1)(cosx -1/2) = 0 where 0≤x<2pia) pi/3, pi/2, 5pi/3b) pi/6, 5pi/6, pic) pi/3, pi, 5pi/3d) pi/6, pi/2, 5pi/6

• sqrt50/sqrt2 (sqrt50)/(sqrt2)*(sqrt2/sqrt2)= (sqrt50*sqrt2)/2 =sqrt100/2=10/2=5 Check my thinking

sqrt50/sqrt2(sqrt50)/(sqrt2)*(sqrt2/sqrt2)= (sqrt50*sqrt2)/2 =sqrt100/2=10/2=5 Check my thinking. thanks... what does this equal: 3sqrt2+4sqrt23 times the square root of 2 PLUS 4 times the square root of 2. This would equal 7 times the square root of 2, or 9.899... THIS IS WAY OV...

• Evaluate cos((7pi)/8) I have this so far cos^2 theta=(1/2)(1+cos(2theta)) cos((7pi/8))= sqroot((1+cos(7pi/4))/2) = sqroot((1+ ((sqroot2)/2))/2) =sqroot((2+sqroot2)/4) =(1/2)(sqroot(2+sqroot2)) but it is marked

Evaluate cos((7pi)/8) I have this so far cos^2 theta=(1/2)(1+cos(2theta)) cos((7pi/8))= sqroot((1+cos(7pi/4))/2) = sqroot((1+ ((sqroot2)/2))/2) =sqroot((2+sqroot2)/4) =(1/2)(sqroot(2+sqroot2)) but it is marked wrong, is there supposed to be a...

• Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya

Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya!translation in english please!

• cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

cos(3π/4+x) + sin (3π/4 -x) = 0= cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx= -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx)I canceled out -1/sqrt2cosx and 1/sqrt2cosxNow I have1/sqrt sinx + 1/sqrt2 sinxAnd that doesn't e...

• If y=cos^2x-sin^2x, then y'= a) -1 b) 0 c) -2(cosx+sinx) d) 2(cosx+sinx) e) -4(cosx)(sinx) I thought the answer was C but the answer key says it is E

If y=cos^2x-sin^2x, then y'=a) -1b) 0c) -2(cosx+sinx)d) 2(cosx+sinx)e) -4(cosx)(sinx)I thought the answer was C but the answer key says it is E. Please help. Thanks in advance.

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