Trig identities and problems??help!!

[fernandotorres]

1.) Simplify: 4 sin 21 cos 21

now i did using one of the identirties 2 sin 2A = 2 sin 2(21)= 2 sin42 but i think im wrong because idk what to do with the 4 because its usually a 2

2.) Prove sin2x + cos x = 0 in intervals of 0 greater than or equal to X < 360

now this is what i did
sin2x + cosx = 0
2sinxcosx + cosx = 0
(cosx)(2sinx + 1) = 0
cosx = 0
sinx = -1/2
cos X's to be 90 and 270 and then for sin X's to be 330 and 210 b/c its neg and sin is negative in quadrant 3 and 4

3.) Prove cos ^2 x = sin^2 x - sin x in the intervals of 0 greater than or equal to X < 360

i dont know how to do this one
^2 means its squared to the power of 2 and x is really that circle with a line across it i know to say it but not spelled it and thank you so much

 

[skeeter]

1.) Simplify: 4 sin 21 cos 21

now i did using one of the identirties 2 sin 2A = 2 sin 2(21)= 2 sin42 but i think im wrong because idk what to do with the 4 because its usually a 2

correct ... 4sin(21)cos(21) = 2[2sin(21)cos(21)] = 2[sin(2*21)] = 2sin(42)

2.) Prove sin2x + cos x = 0 in intervals of 0 greater than or equal to X < 360

now this is what i did
sin2x + cosx = 0
2sinxcosx + cosx = 0
(cosx)(2sinx + 1) = 0
cosx = 0
sinx = -1/2
cos X's to be 90 and 270 and then for sin X's to be 330 and 210 b/c its neg and sin is negative in quadrant 3 and 4

correct ... btw, you're not "proving" anything, you're just solving a conditional trig equation

3.) Prove cos ^2 x = sin^2 x - sin x in the intervals of 0 greater than or equal to X < 360

i dont know how to do this one

change cos[sup:3t4t7xdv]2[/sup:3t4t7xdv]x to 1 - sin[sup:3t4t7xdv]2[/sup:3t4t7xdv]x ...

1 - sin[sup:3t4t7xdv]2[/sup:3t4t7xdv]x = sin[sup:3t4t7xdv]2[/sup:3t4t7xdv]x - sinx

0 = 2sin[sup:3t4t7xdv]2[/sup:3t4t7xdv]x - sinx - 1

now factor and solve like you did in #3

 

[fernandotorres]

thank you so much...you dont how much that means to me....

and number 3 did it and i got

sin delta= -1/2 sin delta = 1
delta= 330, 210 delta=90


thank you again...you really saved my life!!

 

[fernandotorres]

1.) Simplify: 4 sin 21 cos 21

now i did using one of the identirties 2 sin 2A = 2 sin 2(21)= 2 sin42 but i think im wrong because idk what to do with the 4 because its usually a 2

correct ... 4sin(21)cos(21) = 2[2sin(21)cos(21)] = 2[sin(2*21)] = 2sin(42)

2.) Prove sin2x + cos x = 0 in intervals of 0 greater than or equal to X < 360

now this is what i did
sin2x + cosx = 0
2sinxcosx + cosx = 0
(cosx)(2sinx + 1) = 0
cosx = 0
sinx = -1/2
cos X's to be 90 and 270 and then for sin X's to be 330 and 210 b/c its neg and sin is negative in quadrant 3 and 4

correct ... btw, you're not "proving" anything, you're just solving a conditional trig equation

3.) Prove cos ^2 x = sin^2 x - sin x in the intervals of 0 greater than or equal to X < 360

i dont know how to do this one

change cos[sup:3a5dno17]2[/sup:3a5dno17]x to 1 - sin[sup:3a5dno17]2[/sup:3a5dno17]x ...

1 - sin[sup:3a5dno17]2[/sup:3a5dno17]x = sin[sup:3a5dno17]2[/sup:3a5dno17]x - sinx

0 = 2sin[sup:3a5dno17]2[/sup:3a5dno17]x - sinx - 1

now factor and solve like you did in #3

thank you so much...you dont how much that means to me....

and number 3 did it and i got

sin delta= -1/2 sin delta = 1
delta= 330, 210 delta=90


thank you again...you really saved my life!!fernandotorres
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