PART II: WHILE … WEND
SOLUTION:
The answers are in bold, below:
|1) Given Sin θ = [pic], where θ is in the second quadrant, |
|find all the other trig functions using definitions: |
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|First, we draw out a triangle, and fill in the sides we know: [pic] |
|(It's good, but not mandatory to draw it like this, on the x & y axes, so stuff is clearer later on) |
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|We want to use the Pythagorean Theorem, so we write out the formula, and explain it: |
|[pic] |
|a is the line segment between A & C (length is 13) |
|b is the line segment between B & C (length is 5) |
|c is the line segment between A & B (length is not yet known) |
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|Substitute values: [pic] |
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|Square the two numbers: [pic] |
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|Simplify (Subtraction): [pic] |
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|Simplify (Subtraction): [pic] |
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|Square root of each side: [pic] |
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|Simplify (Do the radical) [pic] |
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|Fill in the missing side: [pic] |
|(It's ok to put the missing side onto the same triangle as you drew in step 1) |
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|At this point, you need to either: |
|Write out (in English) a quick explanation about why its actually a negative 12 ("Because the angle is in the second quadrant, and we need to go in the |
|negative X direction in order to get to point C, the 12 is negative"), or else |
|Draw the triangle on an XY coordinate system (clearly labeling the axes), and list 12 as being negative. |
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|Optionally, write out the definitions of the Trig functions. |
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|Since the answer is provided for you, you're done! |
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|Answers: |
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|Sin θ = [pic] |
|Cos θ = [pic] |
|Tan θ = [pic] |
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|Csc θ = [pic] |
|Sec θ = [pic] |
|Cot θ = [pic] |
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|2) Given Sin θ = [pic], where θ is in the second quadrant, |
|with find all the other trig functions using trig identities: |
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|There are 5 different functions you need to find. You can do them in any order, and you don't need to do them exactly like this (i.e., there are multiple, |
|valid, ways you can approach these problems) |
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|CSC: |
|Reciprocal Identity: [pic] |
|Substitute the known value in: [pic] |
|Multiply numerator and denominator by the same value: [pic] |
|Simplify (denominator goes to 1): [pic] |
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|Simplify (denominator of 1 can be ignored) |
|(This step may be combined with the previous one): [pic] |
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|COS: |
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|Pythagorean Identity: Cos2θ + Sin2θ = 1 |
|Substitute the known value in: Cos2θ + [pic] = 1 |
|Do the exponent: Cos2θ + [pic] = 1 |
|Simplify (Subtraction): Cos2θ = 1 - [pic] |
|Simplify (Common denominator, for subtraction): Cos2θ = [pic] |
|Simplify (Subtraction): Cos2θ = [pic] |
|Square root of each side: [pic] |
|Simplify (Do the radical) [pic] |
|Simplify (Do the radical) [pic] |
|Explain (Via picture or English sentence) that being in Quadrant II means that the cosine is negative |
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|SEC: |
|Reciprocal Identity: [pic] |
|Substitute the known value in: [pic] |
|Multiply numerator and denominator by the same value: [pic] |
|Simplify (denominator goes to 1): [pic] |
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|Simplify (denominator of 1 can be ignored) |
|(This step may be combined with the previous one): [pic] |
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|TAN: |
|Quotient Identity: [pic] |
|(We already know Sinθ, Cosθ, so plug those in: [pic] |
|Multiply numerator and denominator by the same value: [pic] |
|Simplify: Denominator cancels to 1, and is ignored, cross-cancel on top: [pic] |
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|COT: |
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|Reciprocal Identity: [pic] |
|Multiply each side by Cotθ, and by [pic], in order to get: [pic] |
|Substitute the known value in: [pic] |
|Multiply numerator and denominator by the same value: [pic] |
|Simplify (denominator goes to 1): [pic] |
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|Simplify (denominator of 1 can be ignored) |
|(This step may be combined with the previous one): [pic] |
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|Answers: |
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|Sin θ = [pic] |
|Cos θ = [pic] |
|Tan θ = [pic] |
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|Csc θ = [pic] |
|Sec θ = [pic] |
|Cot θ = [pic] |
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|3) You're sitting in class one day, and you overhear a classmate, Bob, explaining a problem to someone else. What you hear is the following: "Tangent of 30 |
|degrees? Without using a calculator? No problem. You just draw out your handy-dandy 30-60-90 triangle like this, and notice that the side adjacent to 30° is |
|[pic], the opposite is 1, and the hypotenuse is 2. We could just use the definition of tan, namely that tan = [pic], but where's the fun in that? Let's use |
|the fact that tan = [pic], and that [pic] (which is [pic]), and that cos θ = [pic] (which is [pic] ). Plug those into the equation for tan, get [pic], which |
|simplifies to [pic] - Voila!" List all the mistakes that Bob has made: |
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|(These may be listed in any order) |
|Tangent is [pic] |
|Sine is [pic] |
|Cosine is [pic] |
|Tan is [pic] |
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|Explain why Bob still got the right answer: |
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|Basically, Bob's mistakes w/ Sin & Cos 'canceled out' the mistakes with Tan: |
|Some math might illustrate this better: According to Bob, tan = [pic]. Plugging in what Bob thinks is cos & sin, we get [pic]. Ignore the 1 in the |
|denominator, and multiply top & bottom by ½, and get [pic]. This just happens to be the same values we get if we plug in the correct values of sin ([pic]) and|
|cos ([pic]) into the correct identity for tan ([pic], or [pic]), which then simplifies to [pic], or [pic]. |
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