Posted: 3/18/2014 2:45:19 PM EDT
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It's been about 20 years since I did any trig, but my daughter (no pics) is now in trig. We ran into a problem we couldn't solve last night. And I have vague memories of doing this exact problem in high school. The odd/even identities state that tan(-x) = -tan(x). The same is true for cos, sin, cot, etc. Tan(x) = sin(x)/cos(x). So, tan(-x) = sin(-x)/cos(-x). But then you bring out the negative signs, and you get -sin(x)/-cos(x). And since this is a negative over a negative, you get sin(x)/cos(x), which = tan(x). I know I'm making an incorrect assumption somewhere. I even ran sin(-x)/cos(-x) and -sin(x)/-cos(x) through Wolfram Alpha, and got -tan(x) and tan(x), respectively. And tan(-x) doesn't equal tan(x) and -tan(x) doesn't equal tan(x). Isn't this a classic trig problem? |
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Quoted:
It's been about 20 years since I did any trig, but my daughter (no pics) is now in trig. We ran into a problem we couldn't solve last night. And I have vague memories of doing this exact problem in high school. The odd/even identities state that tan(-x) = -tan(x). The same is true for cos, sin, cot, etc. Tan(x) = sin(x)/cos(x). So, tan(-x) = sin(-x)/cos(-x). But then you bring out the negative signs, and you get -sin(x)/-cos(x). And since this is a negative over a negative, you get sin(x)/cos(x), which = tan(x). I know I'm making an incorrect assumption somewhere. I even ran sin(-x)/cos(-x) and -sin(x)/-cos(x) through Wolfram Alpha, and got -tan(x) and tan(x), respectively. And tan(-x) doesn't equal tan(x) and -tan(x) doesn't equal tan(x). Isn't this a classic trig problem? The error begins in the red part. There are even identities, f(x) = f(-x), and odd identities, -f(x) = f(-x). The -f(x) = f(-x) rule does not apply to all trig functions. Some are odd, some are even. Cosine is an even function, meaning cos(-x) = cos(x). Sine is an odd function, meaning -sin(x)= sin(-x) Therefore, tan(-x) = sin(-x) / cos(-x) = (from the above properties) -sin(x) / cos(x) = -tan(x) |