Trig relations: revision (2 Viewers)

[jane1820]

i thought the answers to c, e, and f are: positive, negative, positive respectfully but the answers are saying: negative, positive, negative respectfully

 

[Study to success]

i thought the answers to c, e, and f are: positive, negative, positive respectfully but the answers are saying: negative, positive, negative respectfully View attachment 46257

I think it's because u have to look at the complementary ratios. For example in C. When sin(90°-θ) it= cos(θ) which is negative in the 2nd quadrant

For e you use supplementary relations so when cos(180°+θ) = -cos(θ) which is the supplement of an acute angle which is So cos in positive.

 

[Average Boreduser]

i thought the answers to c, e, and f are: positive, negative, positive respectfully but the answers are saying: negative, positive, negative respectfully View attachment 46257

it says theta in 2nd quad.

 

[jane1820]

it says theta in 2nd quad.

Yh i see but if u take a for example
cos(180-0)=
if u do 180-A it would give a number between 90 and 0 (assuming A is any angle in the second quadrant)
then this shifts it to the first quadrant where cos is positive
therefore cos(180-0) is positive and not negative despite cos being negative in the second quadrant
(using 0 as theta)

but for c, e and f i dont get it

 

[jane1820]

I think it's because u have to look at the complementary ratios. For example in C. When sin(90°-θ) it= cos(θ) which is negative in the 2nd quadrant

For e you use supplementary relations so when cos(180°+θ) = -cos(θ) which is the supplement of an acute angle which is So cos in positive.

But how about tan(90-0)?
bc according to the rules tan(90-0)=cot(0)
n is how do ik if cotangent is negative?

 

[Average Boreduser]

Yh i see but if u take a for example
cos(180-0)=
if u do 180-A it would give a number between 90 and 0 (assuming A is any angle in the second quadrant)
then this shifts it to the first quadrant where cos is positive
therefore cos(180-0) is positive and not negative despite cos being negative in the second quadrant
(using 0 as theta)

but for c, e and f i dont get it

consider this way. cos(180*+theta)=-costheta. We then sub in any obtuse angle (2nd quad angle). note then that the costheta is also negative. both then cancel out to become positive.

 

[Average Boreduser]

But how about tan(90-0)?
bc according to the rules tan(90-0)=cot(0)
n is how do ik if cotangent is negative?

cotx=1/tanx. sub x which is in 2nd quadrant, makes cotx negative.

 

[pl4smaa21]

I'll try to explain part (e)

 

[jane1820]

consider this way. cos(180*+theta)=-costheta. We then sub in any obtuse angle (2nd quad angle). note then that the costheta is also negative. both then cancel out to become positive.

According to the answers the answer to c is negative not positive
thats what im trying to understand

 

[jane1820]

I'll try to explain part (e)

View attachment 46258

The answer says its negative tho

 

[Average Boreduser]

According to the answers the answer to c is negative not positive
thats what im trying to understand

its incorrect then. expanding out cos(pi+x)=-cosx using compound angle. subbing in values of pi<=x<=pi/2 should result in cosx being negative and thus the answer is going to be positive. Also view from graph its positive in that domain.

 

[pl4smaa21]

so for part (c) i think this is how to do it i think

using complementary angles we see sin(90-theta) = cos(theta)
since theta is in quadrant 2 and the value of cosine is negative in quadrant 2, it must be that sin(90-theta) is negative as it equates to cos(theta) which is always negative if theta is an obtuse angle.


edit: self deprecation is bad

 

[I AM MUSIC]

its incorrect then. expanding out cos(pi+x)=-cosx using compound angle. subbing in values of pi<=x<=pi/2 should result in cosx being negative and thus the answer is going to be positive. Also view from graph its positive in that domain.

no bro wb the hypotenuse cs of the polynomial and ½bh

 

[Average Boreduser]

no bro wb the hypotenuse cs of the polynomial and ½bh

shiiiii I forgot bruh. I might needa pull out the taylor series telecscoping gamma function.

 

[pl4smaa21]

jimmy am I right jimmy??? idk

 

[I AM MUSIC]

shiiiii I forgot bruh. I might needa pull out the taylor series telecscoping gamma function.

allg bbg we all make mistakes

 

[pl4smaa21]

SHIT I'M WRONG man I'M SOO RETARDED

 

[pl4smaa21]

wait fuck i'mright I put in the wrong image damn it

 

[pl4smaa21]

IT'S NEGATIVE YAEEAEAEAEAEAAEEA

 

[Study to success]

so for part (c) i think this is how to do it i think

using complementary angles we see sin(90-theta) = cos(theta)
since theta is in quadrant 2 and the value of cosine is negative in quadrant 2, it must be that sin(90-theta) is negative as it equates to cos(theta) which is always negative if theta is an obtuse angle.


edit: self deprecation is bad

Idk if I’m right but I think: if u substitute an obtuse angle into sin(90-theta) then it will give you a negative active angle. Then if u apply the negative angle unit circle concept then it will be in the 4th quadrant where sin is negative