Does (tanΘ)(cosΘ) always equal sinΘ?
回答 (8)
2016-06-05 7:09 pm
✔ 最佳答案
Yes. Tan(x) = sin(x)/cos(x)So that would be like multiplying (sin(x)/cos(x))*cos(x). The cos(x) will cancel, leaving you with just sin(x). Of course, that is assuming that cos(x) does NOT equal 0.
2016-06-05 7:09 pm
Not always. If cos(t) = 0, then tan(t) * cos(t) is undefined.
2016-06-05 7:23 pm
Yes, except when theta is an odd multiple of 90 degrees (an odd multiple of pi/2 radians). At those points, cos(theta) = 0 and sin(theta) = 1 and tan(theta) is undefined.
2016-06-07 5:55 pm
Yes, since tan theta = sin theta over cos theta
and sin theta over cos theta times cos theta is sin theta
and sin theta over cos theta times cos theta is sin theta
2016-06-05 7:50 pm
Yes, as long as cosΘ is not 0, making tanΘ ± infinity.
2016-06-05 7:22 pm
YES!!!
Because
Tan X = Sin X / CosX
Substituting
(Sin X / Cos X) (Cos X)
Cancel down by 'Cos X'.
Hence
Sin X
Because
Tan X = Sin X / CosX
Substituting
(Sin X / Cos X) (Cos X)
Cancel down by 'Cos X'.
Hence
Sin X
2016-06-05 7:20 pm
Yes
2016-06-05 7:10 pm
yes
2016-06-05 7:08 pm
Well,
the answer is yes
(tanΘ)(cosΘ) = (sinΘ)/(cosΘ) * (cosΘ) = sinΘ
hope it' ll help !!
the answer is yes
(tanΘ)(cosΘ) = (sinΘ)/(cosΘ) * (cosΘ) = sinΘ
hope it' ll help !!
2016-06-05 7:07 pm
Yes.
Think about how they were originally defined using right triangles:
sin() = opp/hyp
cos() = adj/hyp
tan() = opp/adj
So if we have tan() times cos(), we get:
tan() * cos()
(opp/adj) * (adj/hyp)
The two adjacents cancel out leaving:
opp/hyp, which is the definition of sine.
Think about how they were originally defined using right triangles:
sin() = opp/hyp
cos() = adj/hyp
tan() = opp/adj
So if we have tan() times cos(), we get:
tan() * cos()
(opp/adj) * (adj/hyp)
The two adjacents cancel out leaving:
opp/hyp, which is the definition of sine.
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