<p>arccos (2sqrt(5)/5) + arccos (3sqrt(10)/10) = ?????</p>
<p>someone please show me how to work this?</p>
<p>Do you want a new trig identity for that or just solve it? If you want to just solve it, just plug it in your calculator.</p>
<p>I'm not allowed a calculator (it's on a test that doesn't allow calculators)</p>
<p>i'm gonna write arc like (-1) .. arcsinx = sin(-1)x ... right??</p>
<p>let cos(-1)(2sqrt5/5) = x ... and let cos (-1)(3sqrt10/10) = y
we want to find x + y</p>
<p>so cos x = 2sqrt5/5 .. and cos y = 3sqrt10/10</p>
<p>now we want to find x + y .. </p>
<p>cos (x+y) = cosx . cosy - sinx . siny</p>
<p>yeah??</p>
<p>given cos x = 2sqrt5/5 and cos y = 3sqrt10/10 ... use ur triangles to work out sin y and sin x</p>
<p>so sin x = sqrt(25 - (2sqrt5)^2) / 5 = sqrt 5 / 5 </p>
<p>similarly sin y = sqrt10 / 10</p>
<p>therefore cos (x+y) = 3sqrt10/10 . 2sqrt5/5 -- sqrt10 / 10 .sqrt 5 / 5<br>
= 6sqrt50/50 - sqrt50/50
= 5sqrt50/50
= sqrt50/10 = sqrt2 . sqrt 25 / 10 = 5sqrt2 / 10 = sqrt 2 / 2 </p>
<pre><code> = 1 /sqrt 2
</code></pre>
<p>therfore<br>
x + y = pi /4 rad. = 45 degrees</p>
<p>i havn'et checked it .. there might be silly errors but ... thats the method</p>
<p>thanks alot yayaya, that was extremely helpful</p>
<p>you're welcome</p>