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Trigonometry Help


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Note by Dan Ley
1 month, 1 week ago

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Is d equal to a^2-b^2 @Dan Ley Neel Khare · 1 month, 1 week ago

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@Neel Khare I think you can do it by expanding the bottom, and multiplying the top out by \(d+acosc\), and showing that these two expressions are equal. But I can't see a better, nice method. Dan Ley · 1 month, 1 week ago

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@Dan Ley I agree let me work on it I will get back to you after 2-3 days I am a little busy Board exams are coming up Neel Khare · 1 month, 1 week ago

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@Neel Khare That's fine don't stress over it, it's just part of my school project:) Good luck with the exams bro. Dan Ley · 1 month, 1 week ago

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@Dan Ley Thanks But I will try Neel Khare · 1 month, 1 week ago

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@Neel Khare No, \(d^2=a^2-b^2\), so \(d=\sqrt{a^2-b^2}\). Dan Ley · 1 month, 1 week ago

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@Dan Ley Ok Neel Khare · 1 month, 1 week ago

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@Neel Khare don't worry about it, I just multiplied the equation out so that there were no fractions, and replaced \(\sin^2c\) with \(1-\cos^2c\). Then you can show that the two expressions are equal, but it's not fun or elegant or particularly difficult... Dan Ley · 1 month ago

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@Dan Ley I think the right person for this is @Neel Khare Ayush Rai · 1 month, 1 week ago

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