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Need some help on this discussion, I'm not really good at solving trig questions. Pick any expression in the screenshot. Instructions To prove an expression,
Need some help on this discussion, I'm not really good at solving trig questions. Pick any expression in the screenshot.
Instructions To prove an expression, you pick one side of an expression and use the trigonometric identities to rewrite the other side of the expression until it matches the first. It is often helpful to pick the more complicated side, try to convert to sine or cosine, get common denominators or clear fractions, and then use the trigonometric identities. Explain what you did at each step. Here is an example: Prove thatsing + siny coty = cscy 1. Start with the left side, because it is more complicated. 2. Rewrite the cotangent in terms of sine or cosine. 3. sing + siny sin'y cos y = cscy (replaced cotangent with cosine/sine) 4. siny + cos y sing = cscy (sine/sine = 1) 5, sin y siny sing cos y cscy (sine = sine squared divided by sine) 6. sin y +cos y siny cscy (added the fractions because we have a common denominator) 7. 1 siny = cscy (use the first Pythagorean identity) 8. cscy = cscy (use the reciprocal identity) Pick one of the following problems and prove the equivalence similar to the example above. Make sure to explain what you did at each step. Put the function name in the subject line: g(x = sinx cosx tanx = 1 - cos x h(x) = tanx + cot x = secx csca k(x = secx - tanx sina = secx m(x = cscox tax x - 1 = tan 'x For your replies, try to prove their proof again and see if you can solve it a different way. For the example above, another way you could prove might be to: Prove thatsing + siny coty = cscy 1. sing ( 1 + cot y) = cscy(factor out sine) 2. siny (csc y) = cscy(use the third Pythagorean identity) 3. siny 1 sinzy = cscy (use the reciprocal identity) 4. siny = cscy (multiple use fact that sine/sine is 1) 5. cscy = cscy (use reciprocal identity) See the Schedule in the Syllabus and Schedule Module for due dates and the Rubric attached to this Discussion for grading informationStep by Step Solution
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