First, think about how [imath]\sin(A)[/imath] and [imath]\cos(C)[/imath] are related. (That is, first, how are angles A and C related?)
Please show us your attempt at proving the required result (after taking into account the good advice provided above). Then we can then confirm your work as correct or offer further advice on how to proceed.
I hadn't noticed the title.Can someone please explain how do I solve this? How can I find length of AB and DB
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Escape from the corner?When I see the sum of the products of two trig functions I immediately think of....?
I'm curious to know what "help" he might provide other than finding the length of CD (which doesn't seem to enable any progress in proving what's required); he wasn't big on trig, was he? Care to explain?Pythagoras would be able to help you if he was still alive.
@FireBall, it seems that you have managed to confuse all here.
OK, you got me real good!! Good job!!!Escape from the corner?
I'm curious to know what "help" he might provide other than finding the length of CD (which doesn't seem to enable any progress in proving what's required); he wasn't big on trig, was he? Care to explain?
Sometimes it's the best place to be; all the friendliest folks gather there. ???OK, you got me real good!! Good job!!!
I meant think of the "Trig Identity": "sin² θ + cos² θ = 1"
I'll be in the corner for this one for a long time.
I didn't realise anyone here was "confused"... until I read below. ?@FireBall, it seems that you have managed to confuse all here.
That's all good and the fact that they're complementary is the important thing to recognize.In the image you posted it says Prove that : [imath]\sin(A)\cos(C)+\sin(C)\cos\left( \angle D\right)=1[/imath].
[I renamed [imath]\angle EDC~~\angle D[/imath]] Following Pederson's advice, take note that:
[imath]\angle\text{s }A \& C\text{ are complementary}\text{ as are }\angle \text{s }C~ \& ~D [/imath].
Surely, that's another 'redundant' fact; even more confusing (to the OP?) than the suggestion that Pythagoras' theorem is relevant here! The equivalence of those two angles' measures has little bearing on how to prove what's required; it's already clear (from the diagram) that ∠D is the complement of ∠C, there's no need to show it's the same size as ∠A! Sending the OP off down that track is just as much of a wild goose chase as finding side lengths!Does that lead you to the fact that [imath]m(\angle(A))=m(\angle(D))~?[/imath]
The essence of this problem is just the basic Trigonometric Identities! What the OP needs to do is recognize how the Sine of an angle compares to the Cosine of its complement, ie: (for the OP's benefit) how does sin θ° compare to cos (90° - θ°) and vice versa; as (strongly) hinted at in the first two responses. ?Now can you complete the question?
Your answer would be acceptable (apart from the small mistake I have highlighted above) but I suspect that the question writer may have intended that you “prove” the issue without resort to actually entering numerical values. ?Solution:
Since cos C = 4/5 sin C=3/5
since sin A = cos C
therefore sin A= 4/5
Therefore ((4/5)*(4/5)) +((3/5)+(3*5))((3/5)*(3*5))
= (16/28)+(9/25)
=25/25
=1
The answer was right under my nose (sorry?), Thank you very much for your help. I thought the functions get affected by the length of the sides, later I found they remain the same as long as the angle is the same size.