Pythagorean Identities
We know that an identity is always true. So this makes the Pythagorean Theorem an identity, with a^2+b^2=c^2. We replace a,b, and c with x,y, and r making x^2+y^2=r^2. We need to make the theorem equal to 1. We do this by dividing everything by r. Our equation is now (x/r)^2+(y/r)^2=1. From our unit circle we know the ratio of cosine is x/r which means we can substitute cos for x/r. The ratio for sine from the unit circle is y/r and now we can plug in sin for y/r. Plugging both of these in gets us cos^2x+sin^2x=1 which is our main identity.
To derive the identity with secant and tangent from cos^2x+sin^2x=1 we must divide everything by cos^2. We now have sin^2x/cos^2x=1/cos^2x. From the Ratio identity we know tanx=sinx/cosx therefore we are able to plug in tan^2x for sin^2x/cos^2x. We are able to do this because the ratio identities are allowed to power up. As for the other side we know from the Recipricol identity that secx=1/cosx. The Reciprical identity is also allowed to power up, the only one not allowed to is the pythagorean identity. So now we are left with the final identity of 1+tan^2x=sec^2x.
To derive the identity with cosecant and cotangent we must divide everything by sin^2x. Giving us cos^2x/sin^2x+1=1/sin^2x. In the ratio identity cotx=cosx/sinx and even though the problem is squared the ratio identities are allowed to be powered up allowing it to be cot^2x to equal cos^2x/sin^2x. The reciprical identity says that cscx=1/sinx. This identity is also allowed to be powered up, so it equals csc^2x=1/sin^2x. The final equation is cot^2x+1=csc^2x.
INQUIRY ACTIVITY REFLECTION
The connections I see between units N, O, P, and Q are that they all use the trig functions in some way. They are also the same because it is all derived from the unit circle.
If I had to describe trigonometry in three words, they would be, simple once learned.
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