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Math Help - trigonometry (2)

  1. #1
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    trigonometry (2)

    Two lines , at right angle to one another , lie on a plane which is inclined at angle \theta to the horizontal . If the 2 lines intersect at a point on the horizontal and their angles of inclination to the horizontal are \alpha and \beta respectively , prove that

    \sin^2 \theta=\sin^2 \alpha + \sin^2 \beta
    Last edited by mr fantastic; October 7th 2009 at 12:59 AM. Reason: Fixed latex
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Two lines , at right angle to one another , lie on a plane which is inclined at angle \theta to the horizontal . If the 2 lines intersect at a point on the horizontal and their angles of inclination to the horizontal are \alpha and \beta respectively , prove that

    \sin^2 \theta=\sin^2 \alpha + \sin^2 \beta
    Drawing a good 3-D diagram is a great help here, so do the following:

    • Draw a wedge with a horizontal rectangular base PQRS, and a sloping rectangular face PABQ, so that the line AB is parallel to PS, with A vertically above Q and B vertically above R.

      Then the angle that the sloping face makes with the horizontal = \angle APQ = \angle BSR = \theta.


    • Now mark a point O on PS. Join O to A, B, Q, R. Suppose that OA and OB are the two perpendicular lines referred to in the question. Then the angles these lines make with the horizontal are:

      \angle AOQ = \alpha and \angle BOR = \beta

    Then all the triangles are right-angled (except \triangle ROQ). Suppose that AQ = BR = h. Then we have:

    AP = \frac{h}{\sin\theta},\, OA = \frac{h}{\sin\alpha}

    \Rightarrow OP = \sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}

    Similarly OS = \sqrt{\frac{h^2}{\sin^2\beta}-\frac{h^2}{\sin^2\theta}}

    \Rightarrow PS^2 = (PO+OS)^2

    = \Bigg(\sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}+\sqrt{\frac{h^2}{\sin^2\  beta}-\frac{h^2}{\sin^2\theta}}\Bigg)^2

    But PS^2=AB^2=OA^2+OB^2

    \Rightarrow \Bigg(\sqrt{\frac{h^2}{\sin^2\alpha}-\frac{h^2}{\sin^2\theta}}+\sqrt{\frac{h^2}{\sin^2\  beta}-\frac{h^2}{\sin^2\theta}}\Bigg)^2=\frac{h^2}{\sin^  2\alpha}+\frac{h^2}{\sin^2\beta}

    Multiply through by \sin^2\alpha\sin^2\beta\sin^2\theta, and divide by h^2:

    \Big(\sin\beta\sqrt{\sin^2\theta - \sin^2\alpha}+ \sin\alpha\sqrt{\sin^2\theta - \sin^2\beta}\Big)^2=\sin^2\theta(\sin^2\alpha + \sin^2\beta)

    \Rightarrow \sin^2\beta(\sin^2\theta - \sin^2\alpha) + 2\sin\alpha\sin\beta\sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)} +\sin^2\alpha(\sin^2\theta - \sin^2\beta)= \sin^2\theta(\sin^2\alpha + \sin^2\beta)

    \Rightarrow 2\sin\alpha\sin\beta\sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)} - 2\sin^2\alpha\sin^2\beta = 0

    \Rightarrow \sqrt{(\sin^2\theta - \sin^2\alpha)(\sin^2\theta - \sin^2\beta)} =\sin^2\alpha\sin^2\beta , assuming \sin\alpha\sin\beta \ne 0

    \Rightarrow \sin^4\theta-\sin^2\theta(\sin^2\alpha +\sin^2\beta)=0

    \Rightarrow \sin^2\theta = \sin^2\alpha+\sin^2\beta, assuming \sin\theta \ne 0

    Phew!

    Grandad
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